WBCHSE Class 12 Physics Unit 6 Optics Chapter 5 Optical Instruments Notes

Optical Instruments

Optical Instruments Notes:The human eye is a perfect natural optical instrument that is more complex and versatile compared to any man-made optical instrument It may be compared to a camera. If an object is placed in front of a camera, its image is produced in the photographic plate of the camera. Similarly, if an object is placed before a human eye, its image is formed on the screen of the eye Le. retina The different parts of the eye are discussed below

1. Description:

The nearly spherical part of the human eye is called the eyeball. It can move in the eye socket with the help of a few muscles. A brief description of parts of the eyeball which are relevant to our discussion is given below.

Cornea:

It is transparent and its curvature is greater than other portions. Light enters the eye through it Its refractive index is about 1.33

Class 12 Physics Unit 6 Optics Chapter 5 Optical Instruments Structure Of Human Eye

Aqueous humour:

It is a transparent watery liquid. It acts as the refracting medium of light and occupies the space between the cornea and the eye lens. Its refractive index is

Iris- It is a contractile diaphragm with a circular aperture near the center. It is made up of two types of muscles. These two muscles contract the iris and control the intensity of the incoming light. The function of the iris is to adjust and admit a suitable quantity of light to enter the eye through the pupil.

Pupil:

It is a circular aperture at the center of the iris. Through this, light enters the eye. According to the intensity of the incoming light, iris can make the pupil small or large and thus adjust the intensity of the incoming light.

Optical Instruments Notes

Eye lens:

It is a transparent system resembling a double convex lens, being more convex behind than in front. It is suspended behind the iris by some ligaments. The function of the lens is to form real and inverted images of external objects on the retina. The refractive indices of the material of the different pans of the lens are not equal. The average refractive index of the materials is about 1.45.

Suspensory ligament: These ligaments confine the eye lens to the right position.

Ciliary muscle:

These muscles adjust the curvature of eye lens and hence the focal length of the eye lens is changed. By adjusting the focal length of the eye lens we help to form the images of the distant objects unknowingly.

 Vitreous humor:

It is a viscous liquid. It acts as the refracting medium of light and occupies the space between the lens and the retina. Its refractive index is about 1.33

Retina:

It is a semi-transparent sensitive membrane of fibers forming the inner coating of the eyeball. The optic nerves terminate at this membrane and carry the sensation of sight to the brain as soon as an image of any external object Is formed on it

Yellow spot:

In the middle of the rain, there is a portion of diameter 2 mm called the yellow spot. This portion of the retina takes the most effective part in understanding the color and details of the object At the center of the yellow spot there is a circular portion of diameter 0.3 mm. It is known as fovea centralis. This is the most sensitive part of the certain. If The image of an object is formed on fovea centrails it is seen most distinctly.

So the adjusting muscles of the eye always try to cast the image of the fovea centralis. To observe by fovea centralis is on an object on called direct vision and to observe by other portions of the retina is called indirect vision.

Blind Spot:

The least sensitive part of the retina is known as blind spot. If the image of an object is formed at this place it is not visible.

Visual axis and optic axis:

The line joining the center of the cornea and that of the lens is called the optic axis of the eye. The line joining the center of the lens and the fovea centralis is called the visual axis of the eye. The angle between the optic axis and the visual axis ranges from 5° to 7°

2. The function of the eye:

The cornea, aqueous humor, and the lens form a single convex lens with air on one side and vitreous humor on the other. For a normal eye, the focal length of the combination is such that distant objects are focused on the retina.

Rays from external objects enter the eye and undergo a series of refractions at the cornea, aqueous humor, successive layers of the lens, and vitreous humor, and finally form a real, inverted, and diminished image on the retina. Though the image on the retina is inverted the brain intercepts it as If in the erect position.

With the help of the suspensory ligament, the ciliary muscle changes the focal length of the eye lens in such a way that the image of an object is constructed exactly on the retina. So we can clearly observe an object, irrespective of its position

Accommodation of the Eye

For a normal eye, the focal length of the eye lens is such that objects situated at large distances are focused on the retina. In this situation, the value of the focal length of the eye lens is maximum. This focusing is done by the eye by altering the curvature of the lens caused by a change in the tension in the ciliary muscles of the eye lens.

When the eye is in a state of full relaxation, distant objects are focused on the retina. But as the objects towards the eye, the muscles are put to tension to bring back the image on the retina behind the retina

The involuntary process by which the eye adapts its focal length to see objects at all distances is called accommodation. Accommodation power of eye is limited. Due to accommodation, one can see an object clearly up to a distance of 25 cm from his/ her eye. We can also clearly see objects which are very near to us.

Optical Instruments Notes

For example, we can clearly read a book at a small distance from our eyes. But if we bring the book at a distance less than 25 cm from our eyes, then we will not be able to read the book. If we try to read a book in this situation for longer we will feel pain in our eyes.

1. Near point:

It is the nearest position of the object in front of a normal eye up to which the object can be seen distinctly without accommodation. This shortest distance at which an object can be seen distinctly is called the least distance of distinct vision. For a normal eye, this distance is about 25 The process of accommodation cannot function for the normal eye when the object is situated at a distance less than 25 cm from the eye.

2. Far point:

It is the farthest point up to which an object can be distinctly seen without accommodation. For a normal eye the far point is at infinity. The distance between the near point and the far point is called the range of vision. With the help of accommodation, a normal eye can see objects situated within this range of vision.

Adaptation of Eye

It has been said earlier that the circular aperture at the center of the iris is a pupil. With the help of some muscles, Iris can make the pupil small or large and thus adjust the intensity of the incoming light. In the presence of powerful light, the pupil automatically becomes small and in the presence of dim light, it becomes large. The increase or decrease in the size of the pupil according to need is called adaptation.

If the light is suddenly extinguished at night In a lighted do not see anything in the room we room momentarily. The reason is that if light, the pupil of the eye remains small. S owly with time the pupil adapts Itself to the darkness, becomes larger in size and we are able to discern the furniture and other objects in the room. Similarly, in a dark place, the pupil remains large. So when we suddenly come to a lighted place from a dark one, too much light enters the eye at a time and we are unable to see anything

But after some time the pupil automatically becomes small and adjusts the amount of incoming light. So we can see everything distinctly. So, if the intensity of incident light on the eye changes then instantaneously, an adaptation of eye takes place and the above two incidents are observed.

Persistence of Vision

The retina of the eye continues to bear the effect of light after the stimulus has been taken away. This phenomenon is called the persistence of vision. The interval for which the impressions continue is about \(\frac{1}{10}\) of a second. So if two different incidents occur before our eyes within \(\frac{1}{10}\) of a second, we cannot differentiate them. We think that a single incident has happened. This is due to the persistence of vision. For this reason, we cannot separately visualize the blades of a revolving electric fan.

Optical Instruments Notes

In different times blades are in different positions, but it seems that the motion of the blades is continuous. For the same reason, the glowing end of a splinter or match-stick yields a bright circular track when the former is swung around and not bright points changing their positions in a discontinuous manner

In this context, it needs to be mentioned that we can view movies because of the persistence of vision. Numerous still pictures of the same thing taken at short intervals, when moved rapidly within \(\frac{1}{10}\) second before the projector, the discontinuous images fuse together to produce an illusion of continuity

Advantage of two eyes

When we look at an object with our two eyes, we see a single object instead of two. On the retina, two images are indeed formed by the two eye lenses. But the brain blends the two different images into a single one. As the two eyes are located a little distance apart, we see an object from two different sides. With the right eye, we see the front side along with a portion of the right side of the object, similarly with the left eye we see the front side along with a portion of the left side of the object.

This perception of the formation of the two images on the retinas of the two eye lenses creates a single perception in the brain. So we get the three-dimensional idea of the object. The perception of viewing a three-dimensional image of an object with the help of our two eyes is called binocular or stereoscopic vision. We estimate the distance of the different objects from our eyes and understand the actual positions of an object with the help of two eyes. Fitting thread in a needle hole is a difficult task. One eye closed. As it is difficult to visualize the distance of the needle hole and the end of the thread from our eye.

Unit 6 Optics Chapter 5 Optical Instruments Defects Of Vision And Theip Reme – Dies

We know that the range of vision of a normal eye extends from 25 cm from the eye to infinity i.e., if an object is situated anywhere within this long range, its image is formed on the retina. If the range of vision of an eye is less than this normal range the eye is said to be defective.

Optical Instruments Notes

The different defects of vision are stated below:

  1. Long sight or hypermetropia
  2. Short sight or myopia
  3. Presbyopia
  4. Astigmatism

The nature of these defects and the processes of their removal are discussed below

1. Long sight or hypermetropia:

It is the defect of the eye due to which we cannot see the near objects distinctly.

The defect is attributed to either of the two causes:

  1. The eyeball has become too short.
  2. The focal length of the eye lens has become too long.

The power of accommodation of an eye has a certain limit. For a normal eye, the near point is 25 cm. But for a long-sighted eye is greater than 25 cm i.e., objects situated more than 25

Suppose, the point N is the near point of a normal eye. For the defect of long sight, the image of an object situated at N will be formed behind the retina instead of being formed on the retina causing blurring of the image. For him, the near point is situated at N’. So, for this defect, the least distance of distinct vision is more than 25 cm

Class 12 Physics Unit 6 Optics Chapter 5 Optical Instruments Long Sight Or Hypermetropia

Optical Instruments Notes

Remedy:

This defect can be removed by placing a convex lens of suitable focal length in front of the eye. The focal length of the convex lens is such that the light rays starting from the object at N appear to come from N’ after refraction in the lens. So N’ is the virtual image of the object at N.

Class 12 Physics Unit 6 Optics Chapter 5 Optical Instruments Remedy

If D be the least distance of distinct vision for the normal eye and d be the corresponding distance for the long-sighted eye, then the focal length/ of the convex lens placed in front of the eye is given by (from the equation of lens)

⇒ \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

[Here , u = D, v= -d]

⇒ \(\frac{1}{-d}-\frac{1}{-D}=\frac{1}{f} \quad \text { or, } \frac{1}{f}=\frac{1}{25}-\frac{1}{d}\)

Or, f = \(\frac{25 d}{d-25} \mathrm{~cm}\)

[d<25 f is positive ]

The power of that convex lens

= \(\frac{100}{f}=\frac{100}{25 d}(d-25)\)

Or, 4 \(-\frac{100}{d}\)

So long sight can be corrected by using spectacles having convex lenses. As the focal length of convex lens is positive its power is also positive which is why the power of the spectacles is obviously positive.

2. Short sight or myopia:

It is the defect of the eye which; sees near objects1but not the distant objects distinctly.

Optical Instruments Notes

The cause of this defect may be either:

  1. The eyeball has become a little bit elongated or
  2. The focal length of the eye lens has become too short.

For these reasons, the image of a distant object is formed not exactly on the retina, but at a point C in front of it Applying the power of accommodation the image cannot be focussed on the retina So the far point of this short-sighted eye does not remain at infinity but comes nearest to eye i.e, the distance of the air far point of this eye is shorter than that of the normal eye.

Class 12 Physics Unit 6 Optics Chapter 5 Optical Instruments Short Sight Or Myopia

Suppose, F is the far point of a short-sighted eye i.e., if the object is situated at F the eye lens forms its image on the retina. Of course, the near point of this eye remains unchanged, i.e., the least distance of distinct vision of a short-sighted eye remains 25 cm

Class 12 Physics Unit 6 Optics Chapter 5 Optical Instruments Short Sight Eye

So a short-sighted eye does not see distinctly the objects from infinity up to F but can see nearer objects distinctly.

Remedy:

In order to correct the defect, a concave lens of suitable focal length should be used. From this, it is seen that the parallel rays from distant objects are focussed at C in front of the retina.

So to focus the rays on the retina, the point of convergence should be shifted to a further point such that the virtual image of the distant object is brought to the far point F, This is possible when the focal length of the lens is equal to the distance of the far point F, as shown below.

Class 12 Physics Unit 6 Optics Chapter 5 Optical Instruments Remedy.

Optical Instruments Notes

Suppose, the distance of the far point is d and the focal length of the lens L is f. In this case, object distance u = ∞  and image distance v = d

So from equation  \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\), we have

⇒ \(\frac{1}{-d}-\frac{1}{∞}=\frac{1}{f}\)

⇒ \(\frac{1}{f}\) = \(\frac{1}{f}=\frac{1}{-d}\)

Or, f = -d

Optical Instruments Notes

Power of the concave lens.

P= \(\frac{100}{f}\)

P= \(\frac{-100}{d}\) dioptre

So, for remedy of short, the spectacles to be should have a concave lens of focal length equal to the distance of the far point. The focal length of a concave lens is negative. so its power is negative. hence a person having the defect of short sight should use spectacles having negative power.

Sometimes it is difficult to use high-power concave lenses in spectacles. In this case, the far point is bought from infinity to a relatively nearer point of distance d’. If the focal length of lens is f’ in this case, then

Presbyopia:

With the advancement of age, the muscles of the eye lose their elasticity. Hence, the normal power of accommodation of the eye decreases. As a result the least distance of distinct vision increases. So nearer objects are not visible distinctly. This defect is called presbyopia. For remedy of this defect convex less suitable focal length is to be used. Hoxx’exer, far point of the eyre having presbyopia is normal i.e., at infinity.

Again sometimes, die far point of the defective eye comes nearer them infinity. To see distant objects concave lens is to be used. To remove both defects, a convex and a concave lens are used together in a circular framing. Distant objects are seen by the concave lens and nearer objects by the dying convex lens. This type of lens is called a bifocal lens

Again sometimes, die far point of the defective eye comes nearer them infinity. To see distant objects concave lens is to be used. To remove both defects, a convex and a concave lens are used together in a circular framing. Distant objects are seen by the concave lens and nearer objects by the die convex lens This type of lens is called the bifocal lens.

Class 12 Physics Unit 6 Optics Chapter 5 Optical Instruments Biofocal Lens

Astigmatism:

It is an optical defect in which vision becomes blurred along different axes. It is due to the inability of the eye to focus a point object lying in different directions into a sharply focused image on the? retina.

While a normal eye can sec equally distinctly vertical and horizontal lines drawn in a plane an eye having astigmatism does not see all the lines with the same distinctness. A few sets of three parallel lines are drawn inclined at different angles. An eye with astigmatism cannot isolate all of them with the same distinctness.

This defect is due to the unequal curvature of the vertical and horizontal sections of the cornea and is remedied by using cylindrical or sphere cylindrical lenses In order to vary the focal length in one plane. The lenses so used are often called toric lenses.

Optical Instruments Notes

Class 12 Physics Unit 6 Optics Chapter 5 Optical Instruments Toric Lens

Unit 6 Optics Chapter 5 Optical Instruments Defects Of Vision And Theip Reme – Dies Numerical Examples

Example 1. A person having a long slglit cannot see things distinctly at a distance of less than 40 cm. If he wants to see things situated 25 cm from him, what should be the power of his spectacles?
Solution:

Let the focal length of the lens of the spectacles =. Here the power of the spectacles will be such that if an object is situated at a distance of 25 cm its virtual image will be formed at a distance of 40 cm from die eye i.e., u = 25 cm; v = 40 cm; both u and v are negative.

From the equation \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) we have,

⇒ \(\frac{1}{-40}-\frac{1}{-25}=\frac{1}{f} \)

Or, \(\frac{1}{f}=\frac{8-5}{200}=\frac{3}{200}\)

f = \(\frac{200}{3}\) cm

∴ Power of the lens of the spectacles

P = \(\frac{100}{f}\)

Or, \(100 \times \frac{3}{200}\)

= 1.5 m1

= 1.5 D

[Short solution: distance of near point d, = 40 cm

∴ Power of the lens, P= 4 – \(\) = 4 – 2.5 = 1.5 made

As power is positive, the lens is a convex lens.]

Example 2. A short-sighted person can see distinctly the object situated at a distance of 20 cm from him. What type of lens will he use to see the objects situated at a distance of 100 cm from him? What will be the power of the lens?
Solution:

In this case, a lens is to be used to form the image of the objects situated at a distance of 100 cm at 20 cm from the eye i.e., u = -100 cm; v = 20 cm; u and v are both negative

From the equation \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) we have

⇒ \(\frac{1}{-20}-\frac{1}{-100}=\frac{1}{f}\) Or, \(\frac{1}{f}=\frac{-5+1}{100}=-\frac{1}{25}\)

Or, f = -25 cm

So, a concave lens of focal length 25 cm is to be used.

Power of the lens, P = \(\frac{100}{f}=\frac{100}{-25}\)= -4D

Optical Instruments Notes

[Short solution: distance of a far point, d = 20cm

After using the lens distance of the far point, d’ = 100cm

∴ Power of lens, P = \(\frac{100}{d^{\prime}}-\frac{100}{d}=\frac{100}{100}-\frac{100}{20}=-4 \mathrm{D}\)

Since power is negative, the lens is a concave lens

Example 3. A short-sighted person can read a book only up to 15 cm from his eyes. To read a book placed at a distance of 25 cm from him what type of spectacles should he use? What will be the power of the spectacles
Solution:

In this case, a lens is to be used which will form the image of the objects situated at a distance of 25 cm at 15 cm from the man. i.e., u = 25 cm, v = 15 cm, u and v are both positive.

From the equation \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) we have

⇒ \(\frac{1}{15}-\frac{1}{25}=\frac{1}{f}\)

Or, \(\frac{1}{f}=\frac{5-3}{75}=\frac{2}{75}\)

Or, f = \(\frac{75}{2}\)

= 37. 5 cm

So the lens to be used is a concave lens of focal length 37.5

Power of the lens

P = \(\frac{-100}{f}=\frac{-100}{75} \times 2\)

= \(\frac{-8}{3}\)

= – 2.67 dioptre.

Optical Instruments Notes

Example 4. A person can see distinctly up to a distance of 2 no further. To see distinctly up to a long distance what type of spectacles should he use? What will be the of the lens of the spectacles?
Solution:

As the person can see distinctly up to a distance of the defect of the eye is short sight, so to see distinctly up to cave lens. The focal length is 2 m, a long distance he will have to use con of the lens will be such that the image of the objects at infinity will be formed at a distance of 2 m from the eye i.e., u= ∞ ,

v= -2, m = -200 cm (negative)

Suppose the focal length of the lens = f

From the equation \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) we have

⇒ \(\frac{1}{-200}-\frac{1}{\infty}=\frac{1}{f}\)

f = -200cm.

∴ Power of the lens

P = \(\frac{100}{f}=\frac{100}{-200}\)

= 0.5 m

Example 5. A person with spectacles of power 3 m-1 can see distinctly the letters of a newspaper placed at a distance of 25 cm from the eye. At what distance should the newspaper be kept to be able to read It without spectacles
Solution:

The power of the lens of the spectacles = 3 m 1. If f is the focal length of the lens, then

3 = \(\frac{100}{f}\) f = \(\frac{100}{3}\)cm.

Again u = -25cm and f = \(\frac{100}{3}\) cm.

So from the equation

From the equation \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) we have

⇒  \(\frac{1}{v}+\frac{1}{25}=\frac{3}{100}\)

Or, \(\frac{1}{v}=\frac{3}{100}-\frac{1}{25}\)

Or, v = – 100 cm

So to read a paper without spectacles the person will have to place it at a distance of 100 cm from the eye

Example 6. A person can see distinctly any object situated in between the distance 50 cm and 300 cm. What type of spectacles are to be used 

  1. To extend the far point up to infinity and
  2. To bring the least distance of distinct vision in each pair of spectacles

Solution:

To extend the far point from 300 cin to infinity a concave lens is. Since the far point of the defective eye is 300 cm, the focal length of the concave lens of the spectacles will be 300 cm. Because the parallel rays coming from infinity ‘after refraction by the concave lens appear to diverge from a point at a distance of 300 cm from the lens. With these spectacles, the near point of the range of vision will be that object distance for which the image distance is 50 cm.

Here, v = -50 cm and f = -300 cm

From the equation \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) we have

⇒ \(-\frac{1}{50}-\frac{1}{u}=-\frac{1}{300}\)

Or, \(-\frac{1}{u}=\frac{1}{50}-\frac{1}{300}=\frac{1}{60}\)

Or, u= – 60 cm

Optical Instruments Notes

So with this pair of spectacles, the range of vision will be from 60 cm up to infinity.

To bring the near point to 25 cm from 50 cm spectacles with a convex lens are to be used. Here, u = -25 cm and v = -50 cm.

So from the equation \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) we have

⇒ \(\frac{1}{f}=-\frac{1}{50}-\frac{1}{-25}=\frac{1}{50}\)

Or,\(\frac{1}{u}=\frac{1}{-300}-\frac{1}{50}=-\frac{7}{300}\)

To find out the far point v = -300cm, f= 50 cm, u=?

In this case, the range of vision of the man extends from 25 cm to 42.86 cm.

Example 7.  A person using spectacles having power + 2.5 m-1 sees the objects distinctly at a distance of 25 cm. What is the nearest point for the person? What type of defect of vision does the eye have
Solution:

Power of spectacles, P = + 2.5 m-1 If f be the focal length of the lens of the spectacles, then

P = \(\frac{100}{f}\) , Or, f = \(\frac{100}{2.5}\)

= 40 cm

Suppose, the distance of the near point, of, the defective eye = x cm

The lens will form the image of the object situated at a distance of 25 cm at a distance.

Here, u = -25 cm; v m’rx cm and/ = 40 cm

From the equation \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) we have

⇒ \(\frac{1}{-x}-\frac{1}{-25}=\frac{1}{40}\)

Or, \(\frac{1}{x}=\frac{1}{25}-\frac{1}{40}=\frac{3}{200}\)

Or, x = \(\frac{200}{3}\)

= 66.67 cm

This is the distance of the near point for the person i.e., the near point has been shifted away from a normal distance (25 cm). So the defect of the eye is long sight.

Example 8.  A person with defective eyes can see objects distinctly up to a distance of 20 cm. What type of lens? should be used and of what power? From the equation
Solution:

The far point of the defective eyes is 20 cm i.e., under this condition the focal length of the convex lens of the eye, f1 = 20 cm. An external lens of focal length f2 is to be used with the eye lens to shift the far point to infinity. The meaning of the far point at infinity is that the focal length of the combination of the eye lens and external lens, F = oo we have

⇒ \(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}\)

Or, \(\frac{1}{\infty}=\frac{1}{20}+\frac{1}{f_2}\)

f1 = – 20 cm

Since f2 is negative, the external lens is a concave one. its power,

P = \(\frac{100}{f_2}=-\frac{100}{20}\)

Example 9. A boy can clearly see objects between a distance of 15cm to 200cm from his eye. To clearly see an object situated at infinity, what will be the power of the lens that should he use? If he wears that lens then what will be the least distance of distinct vision in that case?
Solution:

To see an object placed at infinity, the focal length of the lens should be such that parallel rays coming from infinity seem to be coming from a point at a distance of 200cm

⇒ \(\frac{1}{-200}-\frac{1}{\infty}=\frac{1}{f}\)

f = – 200 cm

As f is negative, the lens is concave.

Power of the leaf , P = \(\frac{100}{f}\)

= \(\frac{100}{-200}\)

= – 0.5m

Optical Instruments Notes

Let, while he is wearing spectacles, the least distance of distinct vision = u

From the equation \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) we have

⇒ \(\frac{1}{-15}-\frac{1}{u}=\frac{1}{-200}\)

Or, \(\frac{1}{u}=-\frac{1}{15}+\frac{1}{200}\)

= \(\frac{-40+3}{600}=-\frac{37}{600}\)

Or, u = – \(\frac{600}{37}\)

= -16.22 cm.

Hence, the required distance is 16.22

Example 10.  A long-sighted man can clearly see at any distance beyond 2.5m. What kind of lens In his spectacles does require to rent books placed 25cm from his eyes?
Solution:

Let the focal length of the spectacles be f.

Now, if an object is placed 25 cm from the eye then the virtual

The image will be created at a distance of 250cm from the eye

Hence, u = -25cm , v = -2.5m = -250cm

From the equation \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) we get

⇒ \(\frac{1}{-250}-\frac{1}{-25}=\frac{1}{f}\)

Or, f = \(\frac{250}{9}\)

= 27.78.cm

So that person should use the convex lens of focal length 27.78cm.

Unit 6 Optics Chapter 5 Optical Instruments, Optical Instruments, Visual Angle, And Angular Magnification

Optical instruments:

Optical instruments which include magnifying glasses, microscopes, telescopes, etc. are used as aids to vision for clarity and magnification.

Visual, angle:

When the size of the image on the retina, the bigger,is the apparent size of the object to us. The angle subtended by the object at the eye is known as the visual angle. The larger the visual angle, the larger is the size of the image. So the apparent size of the object is determined by the visual angle.

A and B are the two different positions of an object. Suppose the visual angles subtended at the eye are θA and θB at these points. Now, according to the object looks smaller at A than at B. For example, cars and human beings look very small on the roof of a high-rise building. Again if the two objects A and C of different heights subtend equal angles at the eye they Will Appear

Class 12 Physics Unit 6 Optics Chapter 5 Visual Angle

As the object approaches the eye, the visual angle increases. So the apparent size of the object also increases. When the object is placed at the near point the apparent size becomes as great as possible. If the object is nearer to the eye than its near point, the impression on the retina is large but indistinct. So, there is a limit to the distance to which an object can be brought near to the eye. An instrument just removes these shortcomings.

Angular magnification or magnifying power: The magnifying power of a visual instrument is the ratio of the angle subtended at the eye by the image to the angle subtended at the eye by the object. This magnifying power of a visual instrument is called angular magnification.

Angular magnification angle subtended at the eye by the image angle subtended at the eye by the object

It is to be noted that there is a difference between linear magnification (vide the chapter on lens) and the so-called angular magnification in the case of visual instruments. Linear magnification is unnecessary in visual instruments. Because in these instruments instead of the real size of the image, its apparent size is considered example, in a telescope the size of the image of a distant object is much smaller than the real size of the object. In spite of that, the image looks large enough because its visual angle is generally large of the same height.

The size of the sun is much larger than that of the moon. But these two appear of the same size as they subtend equal visual angles at our eyes.

As the object approaches the eye, the visual angle increases. So the apparent size of the object also increases. When the object is placed at the near point the apparent size becomes as great as possible. If the object is nearer to the eye than its near point, the impression on the retina is large but indistinct. So, there is a limit to the distance to which an object can be brought near to the eye. An instrument just removes these shortcomings

Angular magnification or magnifying power

The magnifying power of a visual instrument is the ratio of the angle subtended at the eye by the image to the angle subtended at the eye by the object.

This magnifying power of a visual instrument is called angular magnification.

Optical Instruments Notes

Angular magnification

= \(\frac{\text { angle subtended at eye by the image }}{\text { angle subtended at eye by the object }}\)

It is to be noted that there is a difference between linear magnification (vide the chapter on lens) and the so-called angular magnification in the case of visual instruments. Linear magnification is unnecessary in visual instruments.

Because in these instruments instead of the real size of the image its apparent size is considered e.g., in a telescope the size of the image of a distant object is much smaller than the real size of the object. In spite of that, the image looks large enough because its visual angle is generally large.

Unit 6 Optics Chapter 5 Optical Instruments Microscope

The instrument helps us see very small objects, which are otherwise not visible to the naked eye. It is of two types

  1. A simple microscope or magnifying glass and
  2. Compound microscope.

1. Simple Microscope or Magnifying Glass:

Description and working principle: A simple microscope or a magnifying glass is actually a convex lens of short focal length. We know that if an object is placed within the focal length of a convex lens then an erect, virtual, and magnified image is formed at the same side where the object is placed.

L is a convex lens. Object PQ is placed perpendicular to the principal axis within the focal length of the lens. In this case, a virtual, erect, and magnified image pq is formed which will be seen by an eye placed close behind the lens. The distance of the object from the lens is so adjusted that the image is formed at the least distance of distinct vision (D = 25 cm) from the eye.

Magnification:

In the object, PQ is placed such that its image pq is formed at the least distance of distinct vision. If the lens is thin and the eye is placed very close to the lens then the visual angle β will be given by

β = ∠p1Oq = tan ∠p1Oq = \(\frac{p q}{O q}\)

Since β is very small, tan β = β]

Class 12 Physics Unit 6 Optics Chapter 5 Simple Microscope Or Magnifying Glass

Now, to observe the object distinctly without a lens it must be kept at the least distance of distinct vision i.e., at P1 q. In that case, if the object subtends the visual angle at the eye, then

= ∠pOq = tan ∠pOq = \(\frac{P_1 q}{O q}\)

Angular magnification, the angle subtended by the

m = \(\frac{\text { image placed at the near point }}{\text { angle subtended by the object placed at the near point }}\)

= \(\frac{\beta}{\alpha}=\frac{\frac{p q}{O q}}{\frac{P_1 q}{O q}}=\frac{p q}{P_1 q}\)

= \(\frac{p q}{P Q}=\frac{O q}{O Q}\)

= \(\frac{D}{u}\) = linear magnification of the lens ……………….(1)

[D = least distance of distinct vision or the image distance and u = object distance]

From the equation of the lens, we have,

⇒ \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \text { or, } \frac{D}{-v}-\frac{D}{-u}=\frac{D}{f}\) ……………….(2)

[Here D and v are negative and f is positive]

⇒ \(\frac{D}{u}=\frac{D}{v}+\frac{D}{f}\)

= \(\frac{D}{u}=\frac{D}{v}+\frac{D}{f}\) ……………..(3)

Hence, it is observed that the value of magnification is not constant Magnification depends on the image distance.

If an image is formed at a near point, in that case, v = D

Optical Instruments Notes

Magnification:

m = \(\frac{D}{D}+\frac{D}{f}=1+\frac{D}{f}\) …………….(4)

If an image is formed at infinity, in that case, v = 00

⇒ \(\frac{D}{\infty}+\frac{D}{f}=\frac{D}{f}\) …………….(5)

Maximum value of m = 1+ \(\frac{D}{f}\) and minimum value of m = \(\frac{D}{f}\)

Hence according to the position of image, the magnification lies between (1+\(\frac{D}{f}\) ) and \(\frac{D}{f}\)

For maximum magnification image forms at a near point.

For normal. eye D is equal to 25 cm. In that case,

m = 1+ \(\frac{25}{f}\)

But D is not equal to 25cm for all. So, for different observers, the magnifying power of the same microscope may be different. that magnification may be by diminishing the focal length although not decreasing it much. In that case, the lens will be very thick and the image will be indistinct and distorted. Hence, the magnification of a simple microscope is limited.

Uses:  A magnifying glass is used by

  1. Persons suffering from presbyopia to read small letters,
  2. Biology students to see specimen slides,
  3. Watch repairers to locate defects
  4. Detective department to match fingerprints.

Unit 6 Optics Chapter 5 Optical Instruments Microscope Numerical Examples

Example 1. If an object is placed at a distance of 5 cm from a convex lens, a real image of the object is formed at a distance of 20 cm from the lens. If the lens is used as a magnifying glass what maximum magnification can be obtained from it? Least distance of distinct vision is 24 cm
Solution:

Here u = -5 cm and v = 20 cm.

If f is the focal length of the lens we have

⇒ \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{20}-\frac{1}{-5}=\frac{1}{f}\)

Or, f = 4 cm

So, the focal length of the lens is 4 cm.

Maximum magnification of the magnifying glass

m = \(1+\frac{D}{f}\)

= \(1+\frac{24}{4}\)

= 7

Optical Instruments Notes

Example 2. A watch repairer kept a magnifying glass very close to his eyes and found that the magnifying power of the glass was θ. If the least distance of distinct vision of the eye is 25 cm, calculate the focal length of the lens.
Solution:

Here m = 8 and D = 25 cm

m = \(1+\frac{D}{f}\)

Or, 8 =\(1+\frac{25}{f}\)

f = \(\frac{25}{f}\)

= 3.57 cm

Example 3. A convex lens of power 10 m-1 Is used as a simple magnifying glass. What is the maximum and minimum magnification of the lens?

Power of a lens,

P = \(\frac{100}{f}\) or,= f \(\frac{100}{p}\)

In this case, P = 10 m-1

f  = \(\frac{100}{10}\)  = 10 cm

Maximum magnification = \(1+\frac{D}{f}=1+\frac{25}{10}\)

= 3.5

Maximum magnification =\(\frac{D}{f}=\frac{25}{10}\)

= 2.5

Unit 6 Optics Chapter 5 Optical Instruments Compound Microscope

The magnification produced by a simple microscope is limited. If the object is very small, a simple microscope cannot sufficiently magnify it. For larger magnification compound microscope is to be used. Galileo invented this instrument in the seventeenth century.

1. Description:

In this instrument, two convex lenses of short focal lengths are placed in a tube at a certain distance apart so as to have a common axis. The convex lens O which faces the object to be viewed is known as the objective. The other lens E near the eye is known as the eyepiece. The objective has a smaller focal length and aperture than those of the eyepiece. The tube with the lenses can be moved parallel to the axis towards the object or away from it. There is an arrangement to change the distance between the objective and the eyepiece.

2. Working principle:

Illustrates the working principle of the compound microscope. Fo and Fg are the foci of the objective and the eyepiece.

Class 12 Physics Unit 6 Optics Chapter 5 Optical Instruments Work Principle Of Eyepiece

PQ is a small object. The distance from O is slightly greater than OFo. The objective forms a real, inverted and

Magnified image P1Q1 in front of the eyepiece. The position of the eyepiece Is such that the image P1Q1 formed within its first principal focus Fe. P1Q1 acts as an object to the eyepiece and it forms a magnified virtual image pq of P1Q1.

Optical Instruments Notes

This pq is the final image which is virtual. Inverted and magnified with respect to the object. The distance between the objective and the eyepiece is so adjusted that the final image is formed at the least distance of distinct vision from the eye. This process is called focussing of the microscope. In this case, a highly magnified image is seen without any strain in the eye. The magnification becomes maximum if the final image is formed at the near point of the eye

Magnification:

In the compound microscope magnification takes place in two steps; first by the objective and then by the eyepiece. If the magnification produced by the objective is mo and that produced by the eyepiece is me  then the magnification of the microscope

m = mo × m…………….. (1)

If the distance of the object PQ from the objective is u and the distance of the image Q2 formed by the objective is v, then

mo = \(=\frac{P_1 Q_1}{P Q}=\frac{v}{u}\) ……………(2)

As the eyepiece acts as a magnifying glass and the final image is formed at the near point,

me = \(1+\frac{D}{f_e}\) ……………(3)

[D = least distance of distinct vision and fe = focal length of the eyepiece]

From equations (1), (2) and (3) we have

m = \(\frac{v}{u}\left(1+\frac{D}{f_e}\right)\) ……………(4)

Now, applying the general equation of lens for the objective we have

⇒ \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f_o}\)

fo = (focal length of the objective)

Or, \(1+\frac{v}{u}=\frac{v}{f_o}\)

Or, \(\frac{v}{u}=\frac{v}{f_o}-1\)……………(5)

So magnification of the compound microscope

m = \(\left(\frac{v}{f_o}-1\right)\left(1+\frac{D}{f_e}\right)\) ……………(6)

Since the image P1Q1 of the compound microscope is fanned at the end of the tube, i- = length of the tube =I (approx.)

m = \(\left(\frac{L}{f_o}-1\right)\left(1+\frac{D}{f_e}\right)\) ……………(7)

Since \(f_o \ll L, f_e \ll D\)

So from equation (7) we have,

m = \(\frac{L}{f_o} \cdot \frac{D}{f_e}\)……………(8)

Dependence of magnification on various factors:

Equation (8) indicates that magnification produced by the compound microscope depends on the following factors:

1. Length of die tube (L )

2. Focal length of the objective (fo )

3. Focal length of the eyepiece (fe)

Optical Instruments Notes

However, the length of the tube cannot be increased indiscriminately as it will make handling the instrument difficult. For optimum use and advantage, the length of the rube is maintained within 20-25 cm.

Again, if the magnification of the image is very large, its brightness decreases considerably. So to obtain a sufficiently bright and magnified image the object is to be illuminated with a separate bright light.

To obtain the image free from defects like spherical aberration, chromatic aberration etc. more than one combination of lenses are used for objective and eyepiece.

Uses: In laboratories compound microscopes are widely used for different examinations; for example,  to examine blood cells in medical science, to study different cells for Botany and Zoology, etc

Unit 6 Optics Chapter 5 Optical Instruments Compound Microscope Numerical Examples

Example 1. The focal lengths of the objective and eyepiece of a compound microscope are 0.5cm and 1.5cm respectively. If the least distance of distinct vision is 25cm and magnification is 500 then what is the distance between the objective and the eyepiece?
Solution:

Here fo = 0.5cm, fe = 1.5cm, D =  25cm

Let L  = distance between eyepiece and objective.

We know that the magnification of the compound microscope

m = \(\frac{L D}{f_0 f_e}\)

Or, 500 = \(\frac{L \times 25}{0.5 \times 1.5}\)

Or, L = 15 cm

Optical Instruments Notes

Example 2. The focal lengths of the two lenses of a compound microscope are 0.5 cm and 1 cm respectively. An object is placed at a distance of 1 cm from the objective. If the final image of the object is formed at a distance of 25 cm from the eye, what is the distance between the two lenses and the magnifying power of the microscope?
Solution:

In the compound microscope, the focal length of the eyepiece is greater than that of the objective. So for = 0.5 cm and fe = 1 cm. For the objective u = -1cm.

If  v is the image distance, then according to the equation of the lens

⇒ \(\frac{1}{v}+\frac{1}{1}=\frac{1}{0.5}\)

Or, \(\frac{1}{v}\)

Or, v = 1 cm

Optical Instruments Notes

So, the image is formed on the other side of the objective at a distance of 1 cm.

For the eyepiece, ve = -25 cm (the image is virtual and formed at the near point) and fe = 1 cm

We have from the equation of the lens

⇒ \(-\frac{1}{25}-\frac{1}{u_e}=\frac{1}{1}\)

or, \(\frac{1}{u_e}=\frac{1}{-25}-1\)

= \(-\frac{26}{25}\)

or, \(u_e=-\frac{25}{26}\)

= – 0. 96 cm

The distance between the two lenses

= |v| + |ue| =1 + 0.96 = 1.96 cm

The magnifying power of the microscope,

m = \(\frac{v}{u}\left(1+\frac{D}{f_e}\right)\)

= \(\frac{1}{1}\left(1+\frac{25}{1}\right)\)

= 26

Example 3. The focal lengths of the objective and the eyepiece are 1 cm and 4 cm respectively. The distance between them is 14.5 cm. If an object of height 1 mm is placed at a distance of 1.1 cm from the objective what will be the position and the size of the image seen through the microscope?
Solution:

Here, fo = 1 cm and fe = 4 cm.

For the objective, u = -1.1 cm. If the image of the object is formed at a distance v, we have, from the equation of the lens,

⇒ \(\frac{1}{v}+\frac{1}{1.1}=\frac{1}{1}\)

Or, \(\frac{1}{v}+\frac{10}{11}\) = 1

Or, v = 11cm

So, the image formed by the objective Is at a distance v on the other side of the objective and this image is real. This image acts as an object tor the eyepiece

∴  \(m_1=\frac{v}{u}=\frac{11}{1.1}\) = 10

Now, the object distance relative to the eyepiece

= -(14.5-11) = -3.5 cm

If the image is formed at a distance of V from the eyepiece we have from the equation of the lens

⇒ \(\frac{1}{V}+\frac{1}{3.5}=\frac{1}{4}\)

Or, \(\frac{1}{V}=-\frac{2}{7}+\frac{1}{4}\)

Or, V = -28 cm

This image is virtual and will be formed at a distance of 28 cm in front of the eyepiece

m2 = \(\frac{28}{3.5}\)

= 8

∴ Magnification of the final image

m = m1× m2 × = 10 × 8 = 80

The size of the final image = 80 × 1 = 80 mm = 8 cm

Optical Instruments Notes

Example 4. The focal lengths of the objective and the eyepiece of the compound microscope are 1 cm and 5 cm respectively and the distance between the centres of the lenses is 15 cm. If the final image Is formed at the least distance of distinct vision, what la the magnifying power of the microscope?
Solution:

For the eyepiece, v = least distance of distinct vision =- 25 cm, fe = 5 cm, object distance = u.

According to the equation of the lens, we have,

⇒ \(\frac{1}{-25}-\frac{1}{u}=\frac{1}{5}\)

Or, \(\frac{1}{u}=-\left(\frac{1}{25}+\frac{1}{5}\right)\)

Or, \(-\frac{6}{25}\)

u = – \(\frac{25}{6}\)

So, the image formed by the objective is formed at a distance of \(-\frac{25}{6}\) cm in front of the eyepiece.

Therefore, this image is formed behind the objective at a distance (l5-\(-\frac{25}{6}\) ) or, \(\frac{65}{6}\)cm, The image distance of the object, v = \(-\frac{65}{6}\) cm

Therefore, the total magnification of the compound microscope

m = \(\left(\frac{v}{f_o}-1\right)\left(1+\frac{D}{f_e}\right)\)

= \(\left(\frac{\frac{65}{6}}{1}-1\right)\left(1+\frac{25}{5}\right)\)

= \(\frac{59}{6} \times 6\)

= 59

Example 5. The focal lengths of the objective antiI the eyepiece of n compound microscope arc 1 cm and 2 cm respectively and the distance between them Is 12 cm. If the least distance of distinct vision of the observer Is 25 cm, where should a small object be placed to see it?
Solution:

For the eyepiece, fe = 2 cm, v = – 25 cm. So, from the equation of the lens, we have

⇒ \(\frac{1}{-25}-\frac{1}{u}=\frac{1}{2}\)

Or, \(\frac{1}{u}=-\left(\frac{1}{25}+\frac{1}{2}\right)\)

⇒ \(\frac{27}{50}\)

or, u = – \(\frac{50}{27}\) cm

So, the image formed by the objective is formed in front of the eyepiece at a distance \(\frac{50}{27}\) cm Hence, the image is formed behind the objective at a distance (12- \(\frac{50}{27}\)) or, \(\frac{274}{27}\)cm i.e., the image distance of the object relative to tire objective v = \(\frac{274}{27}\)cm

The focal of the objective fo = 1 cm from the equation of the lens we have,

⇒ \(\frac{1}{\frac{274}{27}}-\frac{1}{u}=\frac{1}{1}\)

Or, \(\frac{27}{274}-\frac{1}{u}\) = 1

Or, \(-\frac{1}{u}=1-\frac{27}{274}\)

= \(\frac{247}{274}\)

Or, u = – \(\frac{274}{247}\) = -1.11 cm

∴ The required object distance = 1.11 cm

Example 6. An object is placed at a distance of 5 cm from the objective of a compound microscope. The final image is formed at the least distance of distinct vision and coincides with the object then calculates the focal lengths of the objective and the eyepiece. Given that the least distance of distinct vision = 25 cm and the magnifying power of the instrument = 15
Solution:

For the objective,

Image distance = v, object distance = u

For the eyepiece, image distance = v1 , object distance = u1

∴ According to the problem,

u = 5 cm and v1 = 25 cm

Optical Instruments Notes

Since the object and the final image coincide, the distance between the objective and the eyepiece is 25-5 = 20 cm.

∴ v +u1 = 20…………………………………………………..(1)

Again, the total magnification of the Instrument magnification by the objective x magnification by the eyepiece.

15 = \(\frac{v}{u} \times \frac{v_1}{u_1}\)

Or, \(\frac{v}{5} \times \frac{25}{u_1}\)

Or, \(\frac{v}{u_1}\)

Solving equations (1) and (2) we have,

v = 15 cm and , = 5 cm

For the objective,

u = -5 cm and v = +15 cm [∴  image is real]

According to the equation of lens we have,

⇒ \(\frac{1}{15}-\frac{1}{-5}=\frac{1}{f_o}\) Or,

fo = 3.75 cm

For the eyepiece,

u1= -5 cm and v1 = -25 cm [as image is virtual]

According to the equation of lens we have,

⇒ \(\frac{1}{-25}-\frac{1}{-5}=\frac{1}{f_e}\)

Or, fe = 6.25

So, the focal lengths of the objective and eyepiece are 3.75 cm and 6.25 cm respectively.

Optical Instruments Notes

Unit 6 Optics Chapter 5 Optical Instruments Telescope

Objects situated at large distances are not visible distinctly with the naked eye as these objects subtend very small visual angles at our eyes. The telescope makes it possible by forming an image that subtends a greater angle as compared to the original object. The image formed is a virtual image.

Telescopes are mainly of two types:

1. Asfronomicaltelescope:

It is used to see astronomical objects such as planets, stars, etc. Here the final image is virtual and inverted relative to the object. Sometimes the astro¬ nomical telescope is also called simply a telescope.

2. Torrosfrlal telescope: It is used to see distant objects situated on the surface or near the surface of the earth here the final image is virtual and erect relative to the object

From the constructional point of view, there Is not much difference between an astronomical telescope and a larger telescope pc. Terrestrial telescope has a provision of erecting the Until image, which astronomical telescope does not have

According to construction, astronomical telescopes are of two types:

  1. Refracting telescope: In this type of telescope, the object tive is made of a single lens or a combination of lenses.
  2. Reflecting telescope: In this type of telescope, a large concave or paraboloidal mirror is used as the objective. Nowadays many different types of telescopes are In use, such as radio telescopes, Infrared telescopes, X-ray telescopes, high energy particle telescopes, gravitational wave telescopes, etc

Refracting Astronomical Telescope:  In this instrument, two convex lenses are mounted ina tube so as to have a common axis. here objective O has an

Class 12 Physics Unit 6 Optics Chapter 5 Refracting Astromical Telescope

Large focal length and a large aperture. Comparatively, eyepiece B has a very small focal length and a very small aperture. The aperture of the eyepiece is taken almost equal to that of the pupil of the eye so that all the refracted rays from the eyepiece enter the eye. The distance between the objective and the eyepiece in the tube can be adjusted by a screw.

Working principle:

Shows that the rays from a distant object are incident on the objective. As the object is situated at a large distance the rays from It may be taken as parallel. So a real, inverted, and very diminished image (BP) compared to the object is formed at the focal plane of the objective. This image acts as an object for the eyepiece which then forms the final image.

Rays from the Image BP diverge and fall on the eyepiece. The eyepiece is so placed that B becomes its first principal focus. Hence, the rays after refraction emerge as parallel rays. To adjust the eyepiece in the proper position is called focusing

After focusing one can easily observe a virtual and largely magnified image through the eyepiece. Since the first image is inverted the virtual image is also inverted with respect to the object. This type of focussing of the telescope is called focussing for infinity and this adjustment is called normal adjustment.

Again, the astronomical telescope may be used in such a way that the final virtual image is formed at the near point of the eye i.e., at the least distance of distinct vision. For this, the eyepiece moved a little distance towards the objective so that the image PQ1 formed by the is the objective comes within the focal length of the eyepiece. Then the eyepiece forms a magnified virtual image pq at the near point of the eye. This image pq is erect with respect to PQ1 but inverted with respect to the original object This type of focusing is called focussing for distinct vision.

Class 12 Physics Unit 6 Optics Chapter 5 Focusssing For Distinct Vision

Optical Instruments Notes

Magnification:

The magnitude of an astronomical or terrestrial telescope Is defined as the ratio of the angle subtended) by the final Image at the eye to the angle subtended by the object. As the object Is situated tit a large distance the angle subtended by the object to the naked eye Is taken to be equal to the angle subtended at the objective. As the eye Is placed very close to the eyepiece the angle subtended by the final image at the eye and the angle subtended at the eyepiece are virtually equal

In the case of focusing on Infinity According to the angular magnification

m = Angle subtended at the eye by the final image Angle subtended at the eye by the object

= \(\frac{\beta}{\alpha}=\frac{\angle F E P}{\angle F O P}=\frac{\tan \angle F E P}{\tan \angle F O P}\)

If is small then tan 0 = 0

m = \(\frac{\frac{F P}{E F}}{\frac{F P}{O F}}=\frac{O F}{E F}=\frac{f_o}{f_e}\)

(fo = focal length of the objective and f = focal length of the eyepiece]

So it is seen that for large magnification, the focal length of the. objective should be large and that of the eyepiece should be small

Again with the help of similar triangles, it can be proved

⇒ \(\frac{f_0}{f_0}=\frac{\text { diameter of the objective }}{\text { diameter of the eye-piece }}\)

m = \(\frac{\text { diameter of the objective }}{\text { diameter of the eye-piece }}\) ………………….(2)

In the case of focusing on distinct vision: According to the angular magnification,

m = \(\frac{\beta}{\alpha}=\frac{\tan \beta}{\tan \alpha}=\frac{\frac{P Q_1}{E Q_1}}{\frac{P Q_1}{O Q_1}}\)

= \(\frac{O Q_1}{E Q_1}=\frac{f_0}{E Q_1}\)………………….(3)

[OQ1 = fo as the objective forms the image PQ j of the distant object at its focal plane.]

Now, PQ1 is the object for refraction in the eyepiece and pq is its image.

Object distance, u = EQ1

As the image pq is formed at the least distance of distinct vision,

Image distance, v =  Eq = D

Therefore, according to the equation of lens

⇒ \(\frac{1}{E q}-\frac{1}{E Q_1}=-\frac{1}{f_e}\)

Or, \(\frac{1}{E Q_1}=\frac{1}{E q}+\frac{1}{f_e}=\frac{1}{D}+\frac{1}{f_e}\)

We get from the equation (3),

m = \(f_o\left(\frac{1}{D}+\frac{1}{f_e}\right)=\frac{f_o}{f_e}\left(\frac{f_e}{D}+1\right)\)………………….(4)

In the case of focusing for infinity as the final image is formed at infinity, so Eq = D →∞

So, from equation (4) we get,

m = \(\frac{f_o}{f_e}\) which supports equation (1)

Optical Instruments Notes

Hence, from equations (1) and (4) it can be observed that in case of focussing for distinct vision, magnification of the telescope increases in the ratio \(\left(\frac{f_c}{D}+1\right)\): 1 If the magnification is very large, the brightness of the image decreases. To increase the brightness of the image necessary arrangement has to be made for allowing light from the object to enter the instrument.

So, the aperture of the objective of this type of telescope is made as large as possible i.e.„ the objective is generally of a large diameter. Moreover, if the aperture of the objective is large, it is possible to examine die object minutely, fe the resolving power of the instrument increases. to remove or reduce the defects of all sorts of aberration, instead of a single lens for each of the objective and the eyepiece a combination of lenses should be used.

Length of the tube of the astronomic telescope:

By the term ‘length of the tube’ of an astronomical telescope, we mean the distance between the objective and the eyepiece. If the length of the tube is L , then

In case of focusing on infinity,

L = fo + fe …………………….. (5)

In case of focussing on a distinct vision

L = OQ1+ EQ1

⇒ \(f_o+\frac{1}{\frac{1}{D}+\frac{1}{f_e}}\)

⇒ \(f_o+\frac{D f_e}{D+f_e}\)…………………….. (5)

Limitations of refracting astronomical telescope: 

As the only lens is used in refracting astronomical telescopes, this instrument has defected to chromatic aberration. For this the final image became hazy. It is a disadvantage to scrutinize the object in views to be noted that, no lens can converge the rays of all in a single point.

For this reason, no image formed becomes distinct, i.e., during the formation of an image, there remains some error all the time. This error is called chromatic aberration.

OH’ To get bright images of distant objects like stars, planets, etc., the aperture of the objective lens is to be made large. However it is extremely difficult and expensive to make an objective lens of a very large aperture. Urns are transparent. That is why, fixing of such a large lens (as an objective) rigidly in the right place is very difficult

Reflecting Astronomical Telescope

In reflecting astronomical telescope (also called a reflector) concave mirror is used in place of the objective lens. It was discovered in the 17th century. It is free from chromatic aberration. Besides, it is easier to construct an objective of a larger aperture and its construction cost is also less.

Mirrors being not transparent, it Is possible to fix a large mirror, as an objective, rigidly in the right place. For this reason, In astronomical studies, most of the telescopes used are of reflecting type. In this regard it is to be noted that the reflecting astronomical telescope is one of the most important instruments used in scientific research

Reflecting astronomical telescopes are of different types, namely the

  1. Gregorian telescope,
  2. Newtonian telescope, and
  3. Cassegrain telescope.

The designs of modern-day telescopes are mainly based on these three types.

Optical Instruments Notes

1. Gregorian telescope: Scottish astrophysicist and mathematician James Gregory published the design of this telescope first in his book in 1663. Based on this, British physicist Robert Hook made the telescope in 1673.

Gregorian Telescope Construction:

The Gregorian telescope consists of two concave mirrors, out of which the primary mirror is parabolic and the secondary one is elliptical in nature. Parallel rays from very long distant objects fall on the primary mirror and getting reflected from it meet at its focusThe secondary mirror is placed beyond the focus of the primary mirror in such a manner that, rays after meeting at focus incident on the secondary mirror as divergent rays.

Then the rays get reflected from it (the secondary mirror) and finally emerge through a small hole made in the primary mirror Thus one can view a magnified erect image through the eyepiece placed at the back of the primary mirror.

Class 12 Physics Unit 6 Optics Chapter 5 Optical Instruments Primary And Secondary Mirror

Gregorian telescope Advantages:

  1. At the focus of the primary mirror, a small but distinct real image of the object under observation is formed.
  2. As a result, the final virtual image becomes much brighter and more distinct

Optical Instruments Notes

2. Newton telescope:

Newton made this telescope in 1668. It was the first usable Gregorian reflector.

Newton Telescope Construction:

In this telescope, objective I is a concave mirror of large focal B length and big aperture. This mirror is placed at one end of a tube of large diameter and the other end of the tube is focused towards distant objects to watch. A plane mirror M2 is placed between the concave lens and its focus. It is inclined at an angle of 15° to the axis of mirror My. The eyepiece E con consists of a convex lens of small focal length and a small aperture is placed in an adjacent tube.

Class 12 Physics Unit 6 Optics Chapter 5 Optical Instruments Newtonian Telescope

Optical Instruments Notes

Newton Telescope Formation of image:

The parallel rays coming from a distant object AB are incident on the mirror Mine. After reflection by’ M2, the reflected converging rays are reflected again by plane mirror M2, and a small real image A’B’ is formed between M2 and E. This image A’B’ acts as an object for eyepiece E. A magnified virtual image A”B” of A’B’ is formed at the least distance of distinct vision or at infinity. If A ‘B’ is on the focus plane of eyepiece E, the final image is formed at infinity

Newton Telescope Magnification: Magnification of telescope

m= \(\frac{\text { angle subtended in eye by the final image }}{\text { angle subtended in eye by the object }}\)

If the final image is formed at infinity, then it can be shown that

m = \(\frac{f_o}{f_e}\)

Here, fo = focal length of concave mirror My which is used as an objective; fe = focal length of eyepiece

Now, fo = \(\frac{R}{2}\) -; R – radius of curvature of the objective. Therefore, the magnification of the reflecting telescope will be increased by increasing the radius of curvature of the concave mirror, used as an objective.

It is to be noted that, distorted images are formed due to rays emerging from the edge of the lens or reflected from the edge of the mirror Hence the images become erroneous. Such errors are called spherical aberration.

To avoid such spherical aberration, a paraboloid mirror is used in a modem reflecting telescope.

Newton Telescope Advantages:

  • The image formed by a reflecting telescope is brighter than the image formed by a refracting telescope.
  • This is because, in a reflecting telescope, no loss of light takes place by reflection and absorption at lens surfaces.
  •  In a reflecting telescope, the use of a parabolic mirror removes spherical aberration and chromatic aberration of the image.
  • The concave mirror used in a reflecting telescope is less expensive than the lens used in a refracting telescope.

Newton Telescope Disadvantages:

  • Alignment of different parts is required every time it is used.
  • There is a central obstruction due to the secondary mirror in the light path, causing the light to be scattered in all directions. Hence, the contrast of the image decreases.
  • At the time of observation of a star through a reflecting telescope, if the eye of the observer is positioned near the edge of the field of view, then stars look like comets. This type of defect Is known as a coma

3. Cassograln reflecting telescope:

Laurent Cassegrain had published the design of this telescope in 1672.

Cassograln telescope Construction: In, a Cassegrain reflecting telescope is

Class 12 Physics Unit 6 Optics Chapter 5 Optical Instruments Cassegrain Reflecting Telescope

In this telescope, the objective is a large parabolic mirror O with a hole in its centre. The aperture of this mirror is large. A small convex hyperbolic mirror M is placed in front of the objective. A convex lens E acts as an eyepiece is placed back of the objective.

In a small-sized telescope, the focal length of the eyepiece is large.

Cassograln telescope Formation of image:

The parallel rays coming from a distant object are incident on the objective O and after reflection from the objective, the rays are incident on the convex mirror M. In the absence of convex mirror M, the reflected rays from the objective would meet at its principal focus.

Hence, in the presence of convex mirror M, the rays coming from the objective are reflected again and a real and inverted image is formed at the point F. This inverted image is seen through the eyepiece E.

In the world, presently many observatories are in function, among which a very notable one is—the Roque de los Muchachos observatory of La Palma, Spain. Its ‘Gran Telescopio Canarias’ is one of the most developed telescopes in the world. Presently it is the largest reflecting telescope with unit aperture in the world. Us effective aperture is 10.4 m and the focal length is 16.5 m. It is situated at the top of a volcano at a height of 2267m from sea level. ”

India can boast of having an observatory at the highest altitude. It is located at a height of 4572 m from sea level at Hanle, Ladakh. Only in this. of the world, the sky remains clear almost 250 days at night

Optical Instruments Notes

Unit 6 Optics Chapter 5 Optical Instruments Numerical Examples

Example 1. The focal lengths of the eyepiece and the objective are 10cm and 200cm respectively. If someone wants to observe the moon with the naked eye through this telescope then what should be the distance between the objective and the eyepiece?
Solution:

If someone wants to observe the moon with the naked eye then the telescope should be focused at the infinity

For focussing at infinity, the length of the tube is

L = fo +fe = 200 + 10 = 210cm

Hence, the distance between the two lenses is 210cm.

Example 2. The length of the tube of an astronomical telescope is 44 cm and its angular magnification is 10. What is the focal length of its objective?

m = \(\frac{f_o}{f_e}\)

Or, fe =  \(\frac{f_o}{m}\)

Length of the tube

L = \(\frac{f_o}{m}=f_o\left(1+\frac{1}{m}\right)\)

L =  \(f_o\left(\frac{m+1}{m}\right)\)

Or, \(f_o=L \frac{m}{m+1}\)

⇒ \(44 \times \frac{10}{10+1}\)

= 40 cm

Example 3. A small astronomical telescope has an objective focal length of 50 cm and an eyepiece of focal length of 5 cm. It is focused on the sun and the final image is formed at a distance of 25 cm from the eyepiece. The diameter of the sun subtends an angle of 32′ at the center of the objective to calculate the angular magnification of the instrument and the actual size of the image.
Solution:

Here, f = 50 cm , f= 5 cm and D = 25 cm .

Angular magnification of the instrument,

m = \(\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)\)

= \(\frac{50}{5}\left(1+\frac{5}{25}\right)\)

= \(10 \times \frac{6}{5}\)

= 12

Optical Instruments Notes

If the angle subtended at the eyepiece by the final image is and the angle subtended at the center of the objective by the sun is a, then

m = \(\frac{\beta}{\alpha}\)

Or β = ma = \(\left(12 \times \frac{32}{60}\right)^{\circ}\)

= \(\frac{32}{5} \times \frac{\pi}{180}\) rad

If the actual size of the image is I, then

Or, β = \(\frac{l}{D}\)

I = βD = \(\frac{32}{5} \times \frac{\pi}{180} \times 25\)

= 2.793

Optical Instruments Notes

Example 4. If the focal lengths of the objective and the eyepiece of an astronomical telescope are 100 cm and 20 cm respectively, calculate the angular magnification of the instrument. If a house of height 60 m situated at a distance of 1 km is observed by the instrument determine the height of the image formed by the objective

Angular magnification of the telescope,

m = \(\frac{f_o}{f_e}=\frac{100}{20}\)

= 5

For the objective

fo = 100 cm , u = -1 km = -105 cm

According to the equation of lens, we have

⇒ \(\frac{1}{v}=-\frac{1}{10^5}+\frac{1}{100}\)

Or, \(\frac{1}{v}=-\frac{1}{10^5}+\frac{1}{100}\)

or, v = 100.1 cm

Magnification \(\frac{v}{u}=\frac{100.1}{10^5}\)

= \(\frac{1}{10^3}\)

= \(=\frac{\text { size of the image }}{\text { size of the object }}\)

Or, \(\frac{1}{10^3}=\frac{\text { size of the image }}{60 \times 100}\)

Size of the image \(\frac{6000}{1000}\)

= 6 cm

Optical Instruments Notes

Example 5. The distance between the Earth and the moon is 386242.56 km and the diameter of the moon is 3218.69km. If the focal length of . the objective of a telescope is 0.018288km then what is the diameter of a real image of the moon formed by the objective?
Solution:

Moon is at a large distance from Earth so the image will be formed at the focal plane of the objective.

Now, \(\frac{\text { distance of the image }}{\text { diameter of the image }}=\frac{\text { distance of the moon }}{\text { diameter of the moon }}\)

Or, \(\frac{0.018288}{\text { diameter of the image }}=\frac{386242.56}{3218.69}\)

∴ Diameter of the image = \(\frac{0.018288}{\text { diameter of the image }}=\frac{386242.56}{3218.69}\)

= 1.52  × 10-4  km

Hence, the diameter of the real image of the moon is 1.52 × 10-4 4km.

Optical Instruments Notes

Example 6. The focal lengths of the objective and the eyepiece of a compound microscope are 2m and 5cm respectively. The distance between the two lenses is 20cm. The final image is formed at a 25cm distance from the objective calculate the distance between the eyepiece and the final image

Here fo = 2m , fe = 5cm

Distance between two lenses (L) = 20cm

Distance between final image and objective (u) = 25cm

The distance between the final image and the eyepiece is

D = l + L = 25 + 20 = 45cm

Unit 6 Optics Chapter 5 Long Question Answers

Question 1. What will happen to the image if one-half of the objective lens is covered with black paper?
Answer:

The size of the image will remain unchanged if one-half of the objective lens is covered with black paper, Only the brightness of the image will decrease to some extent. Here half of the objective lens takes part in the formation of the image. So the complete image is formed. The brightness of the image will be reduced to half as the amount of refracted light is reduced to half

Question 2. The radius of the sun is about 106 km it looks like a disc why?
Answer:

The distance of the Sun from the Earth is about 108 km. So the visual angle subtended at the eyes of a man on the earth is = \(\frac{10^6}{10^8}\). Since this value of the visual angle is very small, the sun looks like a disc

Question 3. Explain how you can identify a telescope and a micro¬ scope from their appearance.
Answer:

The aperture of the objective of the telescope is made large in comparison to its eyepiece to get sufficient light from distant objects. On the other hand, the aperture of the objective of a microscope is small in comparison to its eyepiece. So, by observing the size of the objectives and the eyepieces of the two instruments we can identify them. Moreover, for large magnification, the length of the tube of a telescope is larger than that of the microscope.

Optical Instruments Notes

Question 4. Why is the diameter of the objective of an astronomical telescope made large?
Answer:

As the object is situated at a large distance, it subtends a small visual angle at the objective. So, the image is made large by the astronomical telescope. But in that case, the brightness of the image decreases. If the diameter of the objective is large, a large amount of light from the object enters the telescope and increases the brightness. So, the diameter of the objective of an astronomical telescope is made large

Question 5. Though the lamp posts on a road are of the same height, the distant posts appear shorter—explain the reason.
Answer:

The angle subtended at the eye by the object is called the visual angle. If the visual angle increases, the size of the image formed on the retina also increases. i.e., the object appears to be of large size. The lamp posts situated at large distances from a viewer subtend small angles at the eyes. Hence the lamp posts Seen from large distances appear to be relatively short

Question 6. For making a telescope two lenses of foeat lengths 5 cm and 50 cm are to be used. Which lens will you use for objective
Answer:

Focussing on infinity the magnifying power of a telescope is given by,

m = \(\frac{f_o}{f_c}\)

where fo = focal length of the objective and

fe  = focal length of the eyepiece

So for higher magnifying power should be greater than fe.

Hence the lens of larger focal length i.e., 50 cm is to be used as an objective

Question 7. Which of the following lenses L1, L2, and L3 will you select to construct a best possible 

  1. Telescope,
  2. Microscope

 Which of the selected lenses Is to be used as objective and eyepiece In each case?

Class 12 Physics Unit 6 Optics Chapter 5 Optical Instruments Lens

Optical Instruments Notes

Answer:

A telescope should have an objective of a larger focal length and of larger aperture and an eyepiece of a much smaller focal length and aperture. The power of a lens is less means that the focal length of this lens is large because power is the reciprocal of focal length. So lens Lis to be used as objective (focal length and aperture largest) and L1 as eyepiece (focal length and aperture smallest).

A microscope should have an objective of very small focal length and small aperture and an eyepiece of comparatively larger focal length and larger aperture. So lens L3 is to be used as the objective and  L1 as the eyepiece.

Question 8. What do you mean by resolving power of an optical Instrument? How does the resolving power ora the telescope depend on the wavelength of light and the diameter of the objective lens?
Answer:

The ability of an optical instrument to separate or distinguish close adjacent images is called the resolving power of that optical Instrument.

Resolving power and resolving limit are reciprocal. If the wavelength of light Is λ and the diameter of the objective is d, then the resolving limit of a telescope,

⇒ \(\frac{1.22 \lambda}{d}\)

Resolving power = \(\frac{1}{\theta}=\frac{d}{1.22 \lambda}\)

Hence, if the wavelength of light decreases then resolving power of telescope Increases and if diameter of objective lens is increased then the resolving power of telescope will also increase

Question 9. The focal lengths of the objective and eyepiece of n compound microscope are It mm and 2.5 cm respectively. A man with a normal near point (25 cm ) can focus distinctly on an object placed at a distance of 9 mm from the objective of the microscope. What Is the separation between the lenses and magnification power of the Instrument?
Answer:

For the objective, fo – 0.11 cm and uo = -0.9 cm

From \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) we get,

⇒ \(\frac{0.8 \times(-0.9)}{-0.9+0.8}\) = 7.2 cm

For the eyepiece, fe = 2.5 cm and ve = -25 cm

= \(\frac{v_e f_c}{f_e-v_e}=\frac{-25 \times 2.5}{2.5+25}\)

= \(-\frac{25}{11}\)

= -2.27 cm

Optical Instruments Notes

The distance between the two lenses

vo + ue = 7.2 + 2.27 = 9.47 cm

Magnification power,

m = \(\frac{v_o}{u_o} \times \frac{v_c}{u_e}=\frac{7.2}{0.9} \times \frac{25}{\frac{25}{11}}\)

= 88

Optical Instruments Notes

Question 10.

1. A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, find the angular magnification of the telescope.
Answer: 

2. If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 x 106 m and the radius of the lunar orbit is 3.8 × 108 m
Answer:

Angular magnification \(=\frac{f_o}{f_e}=\frac{15}{0.01}\) = 1500

Since the rays are incident from infinity, the image will be formed at focus.

Diameter of the image

= \(\frac{\text { diameter of moon } \times \text { image distance }}{\text { distance of moon }}\)

= \(\frac{3.48 \times 10^6 \times 15}{3.8 \times 10^8}\)m

= 1.37 cm

Question 11. For a normal eye, the far point is at infinity and the point of vision is about 25 cm. The cornea of the eye provides a converging power of about,40 D and the least converging power of the eye lens behind the cornea is about 20 D. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye lens) of the normal eye.
Answer:

The eye uses its least converging power for an object placed at infinity- Therefore, total converging power of the = 40- 20 = 60 D. Now, for an object at infinity i.e., at u = ∞

Now, for an object at infinity i.e., at u = ∞,

v = f \(\frac{1}{P}=\frac{1}{60} m\)

= \(\frac{5}{3}\) cm

For the image of an object placed at the near point, the focal length is

f = \(\left(\frac{1}{v}-\frac{1}{u}\right)^{-1}=\frac{u v}{u-v}\)

= \(\frac{(-25) \times \frac{5}{3}}{(-25)-\frac{5}{3}}=\frac{25}{16} \mathrm{~cm}\)

Converging power of the lens \(\) = 64 D

Converging power of the eye lens =(64-40) =24D

Thus the range of accommodation of the eye lens is 20 D to 24 D.

Optical Instruments Notes

Question 12. A man with a normal near point (25 cm ) reads a book with a small print using a magnifying glass of focal length 5 cm.

1. What is the closest and farthest distance at which he can read the book when viewing it through the magnifying glass?

2. What is the  maximum and minimum angular magnification  {magnify power) possible using the above simple microscope?
Answer:

Minimum distance of the book from the lens

⇒ \(\frac{f v}{f-v}=-\frac{5 \times 25}{5+25}\)

= -4.2 cm

The maximum distance of the book from the lens is -5 cm because in that case, the image is at ∞

Maximum angular magnification (magnifying power)

⇒ \(\frac{25}{\frac{25}{6}}\) = 6

Minimum angular magnification (magnifying power

⇒  \(\frac{25}{5}\) = 5

Question 13. The magnifying power of a telescope In normal adjustment Is 20, and the focal length of the eyepiece Is 5 × 102 m. What is the magnifying power obtained when (the system Is adjusted so that the final Imago of a distant object Is formed 25 × 102m away from the eyepiece?
Answer:

m = \(\frac{f_0}{f_e}\)

Or, 20 = \(=\frac{f_0}{5 \times 10^{-2}}\)

Or, f = \(f_0=20 \times 5 \times 10^{-2}\)

Now, the magnifying power of the telescope

m = \(f_0\left(\frac{1}{D}+\frac{1}{f_e}\right)=1 \times\left(\frac{1}{25 \times 10^{-2}}+\frac{1}{5 \times 10^{-2}}\right)\)

= 24

Optical Instruments Notes

Question 14. A person who can see objects clearly at n distance of 10cm, requires spectacles to be able to see clearly objects at a distance of 30 cm. What type of spectacle should be used? Find the focal length of the lens

The image of an object at a distance of 50cm has to be formed at a distance of 10cm. In this case, u = -30 cm, v = -10 cm. If the focal length of the lens used Isf then

⇒ \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

⇒ \(\frac{1}{-10}-\frac{1}{-30}\)

⇒ \(\frac{1}{30}-\frac{1}{10}\)

⇒ \(\frac{1-3}{30}=-\frac{1}{15}\)

or,  f = -15 cm

Since f Is negative, the person should use n spectacle with a concave lens

Optical Instruments Notes

Question 15. Define the magnifying power of a compound microscope when the final image is formed at Infinity. Why must both the objective and the eyepiece of a compound microscope hit short focal lengths? Explain.
Answer:

The magnification of compound microscope when the final image is formed at Infinity, m = \(\frac{L}{f_o}\left(\frac{D}{f_e}\right)\) ‘ For producing large magnifying power, both the objective and the eyepiece of a compound microscope should possess short focal length.

When m = \(\frac{L}{f_o}\left(\frac{D}{f_e}\right)\)

f<<L, f<<D,

m = \(\left(\frac{L}{f_o}-1\right)\left(1+\frac{D}{f_e}\right)\)

= \(\approx\left(\frac{L}{f_o} \cdot \frac{D}{f_e}\right)\)

Unit 6 Optics Chapter 5 Short Question And Answers

Question 1. Can we consider spectacles as a visual instrument
Answer:

The instrument which helps us to visualize an object is called a visual instrument. In that sense spectacles; can be called visual instruments. If we have some error in our eyes, we use spectacles to rectify that error and make our vision clearer.  Hence, spectacles can be considered a visual instrument

Question 2. In which telescope the final image Image is erect?
Answer: In terrestrial telescope and (Gallean telescope the final image is erect.

Question 3. What do you moan by binocular vision?
Answer:

Observing a 3-dimensional Image of an object with the help of two eyes Is called binocular vision. So with binocular vision, we can predict not only the length and breadth Inn also the height of an object.

Question 4. What is the main difference between a Galilean telescope and a simple terrestrial telescope?
Answer: 

lf make the final Image erect with respect to the object, the Galilean telescope only has two lenses one Is convex and another one Is concave. For the same reason, a simple terrestrial telescope has three convex lenses

Optical Instruments Notes

Question 5. When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be the short distance between the eye and the eyepiece?
Answer: 

The light rays coming from an object, after refraction through the eyepiece of the compound microscope pass through a small circular region. If we place our eyes on this region, the image is viewed distinctly. This region is the eye ring or exit ring. If we place our eyes very near to the eyepiece, some of the refracted rays would not reach the eye, and the field of view would decrease. Obviously, the right position of the eye ring depends on the distance between the objective and the eyepiece of the microscope

Optical Instruments Notes

Question 6. In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
Answer: 

Yes, there is a very small decrease in the angular magnification. Because the angle subtended by the object at the eye becomes somewhat less than the angle subtended at the lens. However, if the object is at a large distance, then the decrease is negligible.

Question 7. Why should the objective of n telescope have a large focal length and large aperture? Justify your answer.
Answer:

The objective of the telescope is a convex lens of large focal length because it faces distant objects and forms bright Images of distant objects. The aperture of the objective is taken large so that it can gather a sufficient amount of light from distant objects.

 

Unit 6 Optics Chapter 5 Synopsis

If an object is placed in front of the eye then an image of the object is formed on the retina of the eye.

1. Eye lens:

It is a bi-convex lens of variable focal length made of a transparent and flexible substance. The average refractive index for different parts of the lens is approximately 1.45.

2. Accommodation of the eye:

The ability of the eye to alter its focal length and hence to focus the images of objects at various distances on the retina is called accommodation of the eye.

3. The minimum distance from the eye up to which objects can be seen clearly and easily is known as the least distance of distinct vision.

4. The distance between the near and far points of the eye is called the range of vision. Within this range, wherever the object is situated, our eyes can see it by accommodation.

The pupil of our eyes automatically contracts in light of high intensity and expands in light of low intensity. This ability of our eyes is called an adaptation of the eye.

5. When an object produces its image on the retina of the eye, the impression thus produced does not disappear immediately after the object is removed. The impression persists for a second in our brain and it is known as the persistence of vision.

6. When we see an object, each eye forms an image of the object, so that two sets of impressions reach the brain simultaneously and the brain correlates these to get a single impression of the object.

7. As a result, a three-dimensional view of the object can be seen. This kind of vision of a three-dimensional impression of an object by two eyes Is known as binocular vision

8. Different kinds of defects of vision:

Class 12 Physics Unit 6 Optics Chapter 5 Different Kinds Of Defects Of Vision

9. The Instrument by which we can see a magnified image of very small objects is called a microscope.

Optical Instruments Notes

There are two types of microscopes:

  1. Simple microscope or magnifying glass and
  2. Compound microscope.

10. The point at the minimum distance at which an object can be seen easily and clearly by the eye is called the near point of the eye in the case of a normal eye it is about 28 cm from the eye

11. The power lor separating the Images of two objects lying close it, to each other Is called the resolving power of an optical Instrument

12. Angular magnification

= \(\frac{\text { angle subtended by the image at the eve }}{\text { angle subtended by the object at the eye }}\)

Angular magnification in case compound microscope

(where L = length ofthe tube = v +fe.

D = the least distance of distinct vision.

fo = focal length of the objective lens,

fe = focal length of the eyepiece lens]

13. In case of focusing at infinity, the angular magnification of a telescope

m = \(\frac{f_o}{f_e}=\frac{\text { diameter of the objective }}{\text { diameter of the eye-piece }}\)

And the length of the telescope tube. L = fo +fe

In case of focussing for clear vision, angular magnification

m = \(\frac{f_e}{f_e}\left(\frac{f_c}{D}+1\right)\) and length of the telescope tube, L = \(f_o+\frac{D f_e}{D+f_e}\)

Unit 6 Optics Chapter 5 Very Short Question And Answers

Question 1. Give a practical application of the persistence of vision
Answer: Cinema

Question 2. How can the defect of astigmatism be corrected?
Answer: By using spectacles with cylindrical or sphere cylindrical sense

Question 3. The minimum distance of distinct vision for a person is 1 m. What eye defect does he suffer from?
Answer: Long-sightedness

Question 4. If the least distance of distinct vision is D and the focal length of the lens is/ then what is the equation for magnification in a simple microscope?
Answer: \(\left[1+\frac{D}{f}\right]\)

Question 5. If the length of the tube of a compound microscope is increased, the magnification increases—Is this statement true or false?
Answer: True

Optical Instruments Notes

Question 6. In what type of telescope is the final image erect?
Answer: In terrestrial telescope and Galilean telescope

Question 7. What is used for the objective of a reflecting telescope?
Answer: A concave mirror

Question 8. Write the names of two ordinary types of the telescope.
Answer: Astronomical telescope and terrestrial telescope

Question 9. Is the length of the instrument to be changed if the focal length of the objective of an astronomical telescope is increased
Answer: Yes, the length will increase

Unit 6 Optics Chapter 5 Fill In The Blanks

Question 1. The impression of a three-dimensional image created by our two eyes is called___________________
Answer: Binocular vision

Question 2. For a normal eye the least distance of distinct vision is______________
Answer: 25 cm

Question 3. Vision of normal eye range from ____________ to______________________
Answer: 25cm, Infinity

Question 4. ______________ is the eye defect which old people usuallly suffer from ____________________
Answer: Presbyopia

Question 5. Cylindrical lenses are used as a remedy for ________________
Answer: Astigmatism

Question 6. For long – sightedness___________ lens should be used
Answer: Convex

Question 7. For short – sightedness___________ lens should be used
Answer: Concave

Question 8. If the focal length of the microscope is small, magnification is ________________
Answer: Large

Question 9. To increase the magnification of a compound microscope, the objective and the eyepiece of ______________ focal lengths and the microscope tube of ________ length are to be taken
Answer: Small, Large

Question 10. In an astronomical telescope, the focal length and the aperture of the objective are ____________ and the aperture of the eyepiece are taken ____________ compared to those of the objective
Answer: Lrge, Small

Question 11. The final image is an astronomical telescope is_____________ and _____________ with respect to the object
Answer: Virtual, Inverted

Question 12. The final image in a terrestrial telescope is__________ and ________ with respect to the object
Answer: Virtual, erect

Question 13. In Galileo’s telescope, the objective is ______________ lens but the eyepiece is ________________ lens.
Answer: Convex, Concave

Optical Instruments Notes

Unit 6 Optics Chapter 5 Assertion-Reason

Direction:  These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

  1. Statement 1 Is true, statement 2 Is true; statement 2 Is a correct explanation for statement 1
  2. Statement 1 is true, statement It Is true; statement 2 Is not a correct explanation for statement I.
  3. Statement 1 Is true, statement 2 is false.
  4. Statement 1 Is false, statement 2 Is true

Question 1.

Statement 1: An observer looks at a tree of height 15 cm with a telescope of magnifying power 10. To him, the tree appears to be of height 150 m

Statement 2: The magnifying power of a telescope is the ratio of the angle subtended by the image to that subtended by the object.

Answer: 4. Statement 1 Is false, statement 2 Is true

Question 2.

Statement 1: The resolving power of a telescope is greater if the diameter of the objective

Statement 2: An objective lens of large diameter collects more light.

Answer: 1. Statement 1 Is true, statement 2 Is true; statement 2 Is a correct explanation for statement 1

Unit 6 Optics Chapter 5 Optical Instruments Match The Column

Question 1. Match the dependence of column 1 with column 2

Class 12 Physics Unit 6 Optics Chapter 5 Angular Magnification

Answer: 1-B, 2-A,D, 3-C, 4 – B

Optical Instruments Notes

Question 2. Match the defects of the eye in column 1 with their remedies in column 2

Class 12 Physics Unit 6 Optics Chapter 5 Defects Of The Eye And Remedies

Answer: 1-C, 2-D, 3-A, 4-B

Optical Instruments Notes

Question 3. Match the magnification of instruments in column 1 with the corresponding mathematical expressions in column 2

Class 12 Physics Unit 6 Optics Chapter 5

Answer: 1-C, 2-B, 3-A, 4-D

Leave a Comment