WBCHSE Class 12 Physics Reflection Of Light Notes

Reflection Of Light Light Introduction

Any luminous body is a source of light. The sources of light ‘ may be of two kinds

1. Self-luminous source and
2. Nonluminous source.

The sun, the stars, electric bulbs, burning candles, etc. are lumi¬ nous sources. Non-luminous sources themselves become visible when light from a luminous body falls on them.

The moon and the planets are non-luminous bodies. These are visible as light from the sun falls on them and is reflected.

Reflection of light:

When a ray of light passing through a medium is incident on the interface with another medium then, a portion of light returns to the first medium. This phenomenon Is called the Reflection of light

 Reflection, absorption, and refraction of light:

Light travels in a straight line in a homogeneous medium.

When it is traveling in one homogeneous medium and meets the surface of another homogeneous medium.

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The following effects may occur:

1. A portion of the light falling on the surface of separation returns to the first medium. This phenomenon is called the reflection of light.
2. A part of the incident light is absorbed by the second medium.
3.  If the second medium is transparent or translucent then a portion of the incident light penetrates into the second medium and undergoes a change of direction at the surface separating the two media and continues to travel along a straight line.
4. This phenomenon is called refraction of light

The surface from which the reflection of light takes place is called the reflector.

The amount of reflected light depends on the following two factors:

1. Direction of Incident light:  The more obliquely the Inol dent light falls on the reflector, the more the amount of reflected light.
2. Nature of the first and second medium: It Is found from the experiment that if light Is Incident from air to glass nor¬ mally, about 4% of incident light Is reflected, hot If light from air is incident on u plane mirror normally, about W0% of incident light is reflected.

Again if light Is incident from air to glass the amount of light reflected Is more than the amount of light reflected when light Is Incident from air to water. On the other hand, If the reflector is black, most part of the incident light is absorbed by the reflector. As a result, there is negligible reflection from a black body or surface

Reflection Of Light Some Definitions

In M_1, M₂ is a reflector. A ray AO is incident on the reflector at point 0 and is reflected along OB. AO is the incident ray and OB is the reflected ray. 0 is the point of incidence.

The angle of incidence: The angle that the incident ray makes with the normal to the reflector at the point of incidence is called the angle of incidence. ON is the normal drawn on M₁ M₂ at O . Hence, ∠AON is the angle of incidence.

The angle of reflection:

The angle that the reflected ray makes with the normal to the reflector at the point of incidence is called the angle of reflection. ∠BON is the angle of reflection.

Reflection Of Light – Laws Of Reflection

Reflection of light obeys the following two laws:

1. The Incident ray the reflected ray and the normal to the reflecting surface at the point of incidence, are all He on the same plane.
2. The angle of incidence is equal to the angle of reflection.
   ∠AON = ∠BON.

Normal Incidence:

The ray of light AO is incident normally on M₁ M₂ So the angle of incidence is zero. According to the law of reflection, the angle of reflection is also zero. Thus die ray retraces its path into the first medium.

 Reflection Of Light – Types Of Reflection

Reflection can be of two types depending on the nature of the surface of the reflector. O regular reflection and 0 diffused reflection

1. Regular Reflection:

If a parallel beam of rays is incident on a smooth plane reflector, then it is reflected wholly as a parallel beam This type of reflection is called regular reflection. Such reflectors are plane mirrors, upper surfaces of undisturbed water, polished metal surfaces, etc.

In the case of a smooth plane surface, the normals drawn at the points of incidence of the beam of rays are parallel to each other. So in the case of regular reflection, if the incident rays are parallel, the reflected rays are also parallel to each other.

The image of an object is formed due to regular reflection [vide section So, we see our image In a plane mirror

It will be interesting to note that only the part of the reflector that reflects parallel rays into the observer’s eyes appears brighter to the observer than the other part of the reflector.

2. Diffuse Reflection:

A parallel beam of rays after reflection from a rough surface does the Reflection of light obey the following two laws. not remain parallel. Such type of reflection of light is called diffused reflection, of reflection, reflection.

Inoccurs this type on the surface of every visible object, each ray of the incident parallel beam is reflected in its own way without being parallel to one another. This is because

The normals at the points of incidence are not parallel to each other. Hence no image is obtained.

However, it is because of diffuse reflection the objects in our surroundings are visible. When a beam of rays falls on the rough sur¬ faces of those objects, it is scattered in all directions.

So, whatever may be the position of the observer, quite a few rays will invariably enter the eyes. As a result, the observer sees the object more or less destined. In this case the reflector looks almost equally bright from all directions but no image of a source is seen.

Reflection from any plane, curved or rough surface, follows the two laws of reflection.

3. Comparison between Regular and Diffuse Reflections

Reflection Of Light – Deviation Of A Ray Due To Reflection

When a ray of light changes its original course due to reflection. the angle between the original and the final directions of the ray is a measure of the magnitude of deviation of the ray.

It is obvious from the figure that in the absence of the reflector M₁M₂, the M ray AO would have traveled straight in the direction AOC, but has instead been deviated due to reflection. The magnitude of this deviation,

δ = ∠BOC = 180° – ∠AOB = 180°- 2i

Reflection Of Light – Some Phenomena Of Reflection

When light falls on a black body, practically no reflection takes place. The black body absorbs almost the whole incident light. Hence, optical instruments like cameras, tele¬ scopes, etc.

Are painted black on the insides to avoid unwanted reflection. On the other hand, white objects do not absorb any light but rather reflect it. Hence to stop the absorption of light and to increase the brightness, white surface air is used.

This is the reason why white screens are used for projection In the cinema.

Twilight: 

Twilight Is the time between dawn and surface and the time between sunset and dusk. The sun itself is not actually visible from this ground level because It is below the horizon. However, the suspended dust particles in all upper atmospheres still receive direct fuse reflection, this light spreads In all directions and pat daily illuminates the ground daily.

When light falls on a glass slab only a negligible portion Is reflected: most of the rays pass through It. Hence the glass slab Is coated with aluminium1, to make a mirror. As the coating is opaque the Incident rays are almost completely reflected from the silvered surface, with only a small portion of the incident is reflected from the front surface.

Reflection Of Light Image

 Image Definition:

When rays of light diverging from a point source after reflection or refraction converge to or appear to diverge from a second point, the second point Is called the Image of the first point.

When light rays from an see the object at the place whore it Is actually situated. Hut If the rays come after reflection or refraction we see the object elsewhere. What we see In the new position Is actually Its Image

There are two kinds of images:

1. Real linage and
2. Virtual image.

1. Real image:

When rays of light diverging from a point source after reflection or refraction converge to a second point, the second point is called the real Image of the first point.

A real image can be formed on a screen, as evident from, where a convex lens has formed a real image A’ of a point source A.

Some examples of real image:

1. The image formed on the cinema screen
2. The image formed by a camera, etc

2. Virtual image:

When rays of light diverging from a point source after reflection or refraction appear to diverge from a second point, the second point is called the virtual image of the first point.

A virtual image cannot be formed on a screen as illustrated, where a plane mirror has formed a virtual image A’ of the point source A

The image of a tree standing by the side of a pond, on the surface of water is a virtual image. A mirage is also a virtual image.

Differences between real image and virtual image:

Image in a Plane Mirror:

Imags of a point object: Let A be a point source A ray of light AO is incident at O normally and retraces its path along OA.

Another ray AC follows the path CD after reflection. The reflected rays when produced backwards, meet at A’. It appears that the two reflected rays are coming from A’. So, A’ is the virtual image of A. The line AOA’ joins the object and the image is normal to the mirror.

CN is normal at the point of incidence. Since OA and CN
are parallel,

∠OAC = ∠ACN = i (Alternate angles)

and ∠OA’C = ∠NCD = r (Corresponding angles)

But ∠ACN = ∠NCD (i = r)

∠OAC = ∠OA’C

Hence the two right-angled triangles AOC and A’OC are congruent.

∴ OA = OA’

Thus, object distance from the mirror = image distance from the mirror

Hence, the image formed by a plane mirror lies on the perpendicular from the object to the mirror as far behind the mirror as the object is in front of it.

Image of an extended object:

Let AB be an extended object in front of a plane mirror M₁M₂. Every point of the extended object may be regarded as a source of light. The complete image of the object will be obtained by locating the position of the images of all the point sources.

Drawing of the image of an extended object:

From the topmost point A of the extended object perpendicular A01 is drawn and it is extended up to A’ in such a way that A in the previous case, A’ is the image of A. The topmost point of AB.

Similarly, B’ is the image of the lowermost point of AB, For every intermediate point of AJB, a corresponding image will be formed between A’ and B’. So, A’B’ is the image of AB. Obviously AB and A’B’ will be of the same size.

So, a plane mirror forms a virtual image of the same size of an extended object.

It is also obvious from the figure that the eye can catch the image within the portion PS of the mirror, as long as the relative positions of the eye and the mirror are not changed

Lateral insertion Definition:

An image of an object formed in a plane mirror is inverted sideways. This effect of plane mirrors is called lateral inversion.

The letter P held in front of a plane mirror will be seen as laterally inverted. The lateral turning is due to the fact, that every point image is at the same distance behind the plane mirror B as the point object is in front of it,

Being the size of the image equal to the size of the object. If the mirror is held vertically, it does not. invert the image which means turning an image upside down. Conversely, if a point source is placed at the point / its virtual image is formed at the point O.

This is due to the principle of reversibility of light rays. Only the image is laterally inverted. If we move our’ right hand facing a mirror, we see the image moving its left hand.

The perpendicular distance of every point of the object from the plane mirror = the perpendicular distance of every image point from the mirror.

i.e., AO₁ =  A’O₁; BO₂ = B’O₂.

So, the image is laterally inverted.

The images, due to bodies having symmetrical sides such as a sphere or letters like A, H, M, I, O, T, etc. are not affected by lateral inversion, but the images with non-symmetrical bodies like mug or letters like P, B, C, etc. are affected.

The characteristics of the image formed by a plane mirror:

1. With respect to the mirror, object distance = image distance.
2. The straight line joining the object and its image is perpendicular to the mirror.
3. The image is virtual.
4.  The image is formed behind the mirror and is the same size as that of the object.
5. The image is laterally inverted.

Image in a Plane Mirror due to a Conver- gent Beam:

Suppose a converging beam of light is incident on a plane mirror M₁M₂. In the absence of the mirror, the rays would converge. Due to reflection, the rays PQ and RS meet at point I instead of O. It can be proved that the straight line joining I and O is perpendicular to the mirror and IA = OA. Point O is called the virtual object and point I is its real image. Thus a plane can form a real image of a virtual object

Conversely, if a point source is placed at point I its virtual image is formed at point O. This is due to the principle of reversibility of light rays.

For a given incident ray if the mirror is rotated through an angle, the reflected ray turns through an angle of 2θ.

The minimum length of the plane mirror required to have the full-length image of a person standing in front of it is equal to half the height h of the person i.e. \(\frac{h}{2}\)

If an object moves towards (or away from) a plane mirror with a speed v, the image of the object will move towards (or away from) the mirror with a speed 2v.

4. If two plane mirrors facing each other are inclined at an angle with each other, the number of images formed due to multiple reflections, is given by n = 360°-1. If (36-1) is not an integer, the next integer will indicate the number of images.
360°

5. Due to reflection, the frequency, wavelength and speed of light are not changed.

6. The intensity of light after reflection decreases.

7. For reflection of light from a denser medium a phase change of occurs.

Reflection Of Light – Curved Reflecting Surface

Mirrors used in torches, headlights, or viewfinders of vehicles are curved mirrors. Curved mirrors may be spherical, cylindrical, or parabolic. In this chapter, however, we shall restrict our discussion only to spherical mirrors.

The laws of regular reflection are equally applicable to curved surfaces as well. But in this case, the position of the image and its size differ widely, as we shall see later.

Reflection Of Light Spherical Mirror

A spherical mirror is a part of a hollow sphere or a spherical surface.

There are two types of spherical mirrors:

1. Concave mirror and
2. Convex mirror.

When the inner surface of a spherical mirror acts as a reflector, it is a concave mirror.  When the outer surface of a spherical mirror acts as a reflector, it is a convex mirror.

Some Related Terms:

Pole:

The center of the spherical reflecting surface is called the pole of the mirror. In  O is the pole.

Centre of curvature:

The center of the sphere of which the spherical mirror is a part is called the center of curvature of the mirror. C is the center of curvature of the mirror MOM’. Obviously, the center of curvature of the concave mirror is in front of the reflecting surface while in the case of the convex mirror it is behind the reflecting surface.

The radius of curvature:

It is the radius of that sphere of which the mirror is a part.  OC is the radius of curvature.

Principal axis:

The line passing through the center of curvature and the pole of the mirror is called the principal axis of the mirror.  XX’ is the principal axis.

Aperture:

The line joining the two extreme points on the periphery of a spherical mirror is called the aperture of the mirror. The angle subtended at the center of curvature by the line is called the angular aperture of the mirror. In the line MM’ is the aperture of the spherical mirror and ZMCM’ is its angular aperture.

The discussion in this chapter will be confined to spherical mirrors of small apertures not exceeding 10°, although, for the sake of clarity, illustrative diagrams will indicate larger apertures.

Paraxial and non-paraxial or marginal rays:

1. Paraxial rays:

Rays that are incident very close to the pole and form a very small angle with the principal axis of a spherical mirror are called paraxial rays.

2. Non-paraxial or marginal rays:

Rays that are incident very far away from the pole or near the margin of a spherical mirror and form a comparatively large angle with the principal axis are called non-paraxial or marginal rays.

All rays incident on a spherical mirror whose aperture is negligibly small compared to its radius of curvature, are considered to be paraxial rays. For further discussions, we will assume all spherical mirrors to be of a small aperture and a comparatively large radius of curvature.

Principal focus:

If rays of light parallel to the principal axis are incident on a spherical mirror, the rays after reflection from the mirror converge to a point on the principal axis in case of a concave mirror and appear to diverge from a point on the principal axis behind the mirror in the case of a convex mirror, this point is called the principal focus or simply focus of the mirror.

F is the principal focus of M the concave mirror and convex mirror respectively.  So, the focus of the concave mirror is real and that of the convex mirror is virtual.

It can alternatively be defined as the point on the principal axis of a spherical mirror at which the image of an object placed at infinity is formed.

According to the principle of reversibility of light rays, it can be said:

The rays diverging from the principal focus of a concave mirror proceed parallel to the principal axis after reflection from the mirror

The rays appearing to converge to the principal focus of a convex mirror proceed parallel to the principal axis after reflection from the mirror.

Focal length:

The distance between the principal focus and the pole of the mirror is called the focal length of the mirror. OF is the focal length. The focal length of a spherical mirror does not depend on the color of the incident light.

Focal plane and secondary focus:

The focal plane of a spherical mirror is the imaginary plane passing through the principal focus at right angles to the principal axis of the mirror.

If a parallel beam of rays is incident on a spherical mirror such that it is inclined to the principal axis then, after reflection the reflected rays converge to the point. on the focal plane in the case of a concave mirror and appear to diverge from the point. On the focal plane in case of a convex mirror. The point F₁ is called the secondary focus of the spherical

The principal focus is a fixed point on the principal axis. However, the secondary focus is not a fixed point. If the angle of inclination of the parallel rays with the principal axis is changed, the position of the secondary focus also changes. But a secondary focus always lies on the focal plane.

Relation between Focal Length and Radius of Curvature

1. In the case of the concave mirror:

Let MOM’ be a concave mirror of a small aperture. C, F, and O are the center of curvature, focus, and pole of the mirror respectively. Ray PQ is parallel to the principal axis, hence passing through F after reflection, CQ being the radius of curvature is perpendicular to the mirror at Q.

∴ ∠PQC = ∠FQC

∴  ∠PQC = ∠QCF [alternate angles]

∠FQC = ∠QCF i.e., Δ QCF is an isosceles triangle.

Hence, FQ = FC

Since the aperture of the mirror is very small, Q and O are very close to each other. So, FQ = FO.

∴ FO = FC or, FO = \(\frac{1}{2}\) OC

i.e f= \(\frac{r}{c}\), where f is focal length and r the radius of curvature.

2. In case of convex mirror:

Let MOM₁, be a convex mirror of a small aperture. C, F, and O are the center of curvature, focus, and pole of the mirror respectively. OC is the radius of curvature.

A ray PQ parallel to the principal axis is incident at Q of the mirror. After reflection, the ray QR appears to come from F. The points C and Q are joined and is extended to N. Since the CQ = CO radius of curvature of the mirror, QN is normal at incidence point Q on the mirror.

If a parallel beam of rays is incident on a spherical mirror such that it is inclined to the principal axis then, after reflection, the reflected rays converge to point F, on the focal plane in case of a concave mirror and appear to diverge from the point F,  on the focal plane in case of a convex mir- But r. The point F is called the secondary focus of the spherical

∠PQN = ∠RQN

∠RQN = ∠CQF [vertically opposite]

∴ ∠PQN = ∠CQF

∴  Since PQ and OC are parallel.

∴ ∠PQN = ∠FCQ [corresponding angles]

∴ ∠FCQ = ∠CQF

Hence , FQ = FC

Since the aperture of the mirror is small, Q and O are very 1. Object at infinity: If an object is at infinity, the rays are close to each other.

So, FQ = FO.

∴ FO = FC

Or, FO = \(\frac{1}{2}\) OC

I.e f = \(\frac{r}{2}\)

Hence the focal length of a spherical mirror is small Thus, the image of an object situ- aperture is equal to half of its radius of curvature.

Reflecting power:

Reflecting the power of a spherical mirror, D = \(\frac{1}{f}\) =\(\frac{2}{r}\) = As both focal length and radius of curvature of a plane mirror are infinite, so power of a plane mirror is zero.

Reflection Of Light – Image Formation Of Extended Object By Spherical Mirror

Ray tracing method:

The position, nature, and size of the image of an extended object, formed by a spherical mirror can be determined geometrically. Any extended object can be considered as the sum of point objects and the images of the point objects constitute the image of the extended object.

Any two of the following rays intersecting at a point will indicate the portion of the image:

1. A ray parallel to the principal axis: After reflection passes through the focus, or appears to diverge from the focus.
2. A ray passes through the focus: After reflection emerges In the case of a concave mirror if the object is placed at the center parallel to the principal axis.
3. A ray passing through the center of curvature: After reflection retraces its path in the opposite direction.

Because the ray passing through the center of curvature incidence on the mirror is the normal incidence in this case.

Image Formation by Convex Mirror:

Objectivity at infinity:

If an object is at infinity, the rays coming from it may be assumed to be parallel and hence after reflection will meet at the principal focus F. If the rays are A oblique after reflection they will meet at a secondary focus F’

Thus, the image of an object situated at infinity is formed on the focal plane. The image is real, inverted, and very much diminished in size

2. Object placed between focus and center of curvature:

An object PQ is placed beyond C on the principal axis of the concave mirror MOM’. A ray PA starting Anotherray PB passing through C is reflected back along BP. The two reflected rays AF and BP meet at p.

So, the real image of P is formed at p. Normal pq is drawn on the principal axis. PQ is the image of PQ. The image is situated between F and C. The image is real, inverted, and diminished relative to the object.

3. Object at the center of curvature :

An object PQ stands at C A ray PA parallel to the principal axis is reflected through F along AF. Another ray PB through F is reflected along BD parallel to the principal axis. The two reflected rays AF and BD meet at p. So, the real image. of P is formed at p. pq is D drawn normally on the principal axis. pq is the image of PQ.

The image is real, inverted, and magnified, and of the same size as the object, and formed at the center of curvature itself.

In case of a concave mirror if the object is placed at the center of curvature the image is also formed at the center of curvature. Hence only in this case, the intervening distance between the object and image is minimum and is equal to zero

4. Object placed between focus and centre of curvature

An object PQ stands between F and C. A ray PA parallel to the principal axis is reflected through F along AF. Another ray PB incident normally at B goes back along BC. These two reflected rays meet at p. So, the image of P is formed at p. pq is drawn normally on the principal axis. pq is the image of PQ.

5. Object at focus:

An object PQ is placed at the focus F . A ray PA parallel to the principal axis is reflected through F along AF. Another ray PB incident normally at B goes back along BC. These two reflected rays being parallel to each other meet at infinity producing the image of P.

6. Object placed between focus and pole:

An object PQ is placed between the pole and focus. A ray PA parallel to the principal axis is reflected through F along AF. Another ray PB incident normally at B goes back along BC. These two reflected raj’s which are divergent would not meet anywhere. But when they are produced backwards they meet at p. Thus they appear to diverge from p. pq is drawn normally on the principal axis. Thus pq is the image of PQ

The Image is virtual, erect, magnified, and situated behind the mirror

 Image Formation by Convex Mirror:

Consider an object PQ in front of a convex mirror MOM. A ray. PA parallel to the principal axis goes back along AD. Ray AD appears to come from the focus. Another ray PB incident normally at B is reflected along BP. Reflected rays AD and BP, produced backwards appear to come from p. So. p is the virtual image of P. pq is drawn normally on the principal axis. Thus pq is the image of PQ.

The image is virtual, erect, diminished in size, and situated behind the mirror.

For any portion of the object in front of a convex mirror, the image is real, inverted, and magnified infinitely, and situation the image will always be formed behind the mirror. This image is rated at infinity.

Comparative Study of Real and Virtual Images in the Case of Spherical Mirror

1. Characteristics of real image:

• It is formed on the same side of the mirror as the object. It is always inverted.
• Real image is not formed in a convex mirror

2. Characteristics of virtual image:

• It is always formed on the opposite side of the mirror as the object.
• It is always erect.
• The size of the virtual image becomes larger than the object or equal to it in the case of a concave mirror. Whereas in the case of a convex mirror, it is smaller than the object or equal to it.

Sign Convention for Spherical Mirror

1.  Cartesian sign convention:

• All distances are to be measured from the pole of the spherical mirror.
• All distances measured in a direction opposite to that of the incident rays are to be taken as negative and all distances measured in the same direction as that of the incident rays are to be taken as positive.
• If the principal axis of the mirror is taken as the x-axis, the upward distance along the positive y-axis is taken as positive while the downward distance along the negative y-axis is taken as negative.

See the concave and convex mirrors given in below to understand the above rules.

The nature sign of object distance(u), image distance(v) focal lengh(f), radius of curvature (r), and height of the image in case of the image formation of a real object by a spherical mirror is given in the following table.

Reflection Of Light – Relation Among Object Distance, Image Distance, And Focal Length

1. In case of concave mirror:

Let O, F, C, and OQ be the pole, focus, center of curvature, and principal axis of a concave mirror MOM’ respectively. PQ is an object placed perpendicularly on the principal axis in front of the mirror. A ray PA, parallel to the principal axis, after reflection, does not form a real image that passes through F.

Another ray PA’, passing through C, after reflection goes back following the same path. These two reflected rays cut each other at p. Hence p is the real image of P. pq is the image of PQ. AB is drawn perpendicular to the principal axis.

Triangles PCQ and pCq are similar

∴ \(\frac{P Q}{p q}=\frac{C Q}{C q}\)

Again,  Δ ABF and Δ  pqF are similar

∴ \(\frac{A B}{p q}=\frac{B F}{F q}\)

Or,  \(\frac{P Q}{p q}=\frac{B F}{F q}\)

[ AB = PQ]

From (1) and (2) we get \(\frac{C Q}{C q}=\frac{B F}{F q}\)

Since the mirror is of small aperture It can be assumed BF ≈ OF

∴ \(\frac{C Q}{C q}=\frac{O F}{F q}\)

Object distance, OQ = -u;

Image distance

Oq = -v; focal

Length, OF = -f

Radius of curvature, OC = -r = -2f

∴  CQ = OQ- OC  = -u+ 2f

Cq = OC- Oq = -2f+v

Fq = Oq- OF = -v+f

From (3) we get, \(\frac{-u+2 f}{-2 f+v}=\frac{-f}{-v+f}\)

or, uv- uf- 2fv² + 2f² = fv

or, uv = uf+ vf

Dividing both sides by uvf we get

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

Or, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}=\frac{2}{r}\)

2. In the case of the convex mirror:

Let O, F, C, and CQ be the pole, focus, center of curvature, and principal axis of a convex mirror MOM’ respectively [Fig. 1.28]. PQ is an object placed perpendicularly on the principal axis in front of the mirror. A ray PA parallel to the principal axis and another ray PD proceed- ing to the centre of curvature, form a virtual image p of P after reflection from the mirror. pq is the image of PQ. AB is drawn perpendicular on the principal axis.

Triangles PCQ and pCq are similar

∴ \(\frac{P Q}{p q}=\frac{C Q}{C q}\)

Again ΔABF and ΔpqF are similar

∴ \(\frac{A B}{p q}=\frac{B F}{F q}\)

Or  \(\frac{P Q}{p q}=\frac{B F}{F q}\)

From (5) and (6) we get,

∴ \(\frac{C Q}{C q}=\frac{B F}{F q}\)

Since the mirror is of small aperture It can be assumed BF ≈ OF

∴ \(\frac{C Q}{C q}=\frac{O F}{F q}\)

Object distance, OQ = -u;

Image distance, Oq = +v; focal

Length, OF = +f

Radius of curvature, OC = +r = +2f

∴  CQ = OQ +OC  = -u+ 2f

Cq = OC- Oq = +2f+v

Fq = OF- Oq = +f – v

From (7) we get,

⇒ \(\frac{-u+2 f}{+2 f+v}=\frac{+f}{+f-v}\)

or, 2f²- vf= – uf+2f²+uv-2vf

or, uv = uf+ vf

Dividing both sides by uvf we get

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

Or, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}=\frac{2}{r}\)

It may be noticed that using the same sign convention, the relation between the various distances is the same for both concave mirror and convex mirror. This relation viz., += = 2 is called the mirror equation or spherical mirror equation.

In case of a concave mirror if u and v are the object distance and the image distance of a real object and its real image respectively, then the u-v graph is a rectangular hyperbola and the graph of their reciprocals ( \(\frac{1}{u}\) – \(\frac{1}{v}\) graph) is a straight line.

Equation of plane mirror from the equation of spherical mirror:

The mirror equation is

⇒ \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)= \(\frac{2}{r}\)

For a plane mirror, r→ ∞

∴ \(\frac{1}{v}\) +\(\frac{1}{u}\) = 0

Or, v= -u

This proves that in the case of a plane mirror, the image is as far behind the mirror as the object is in front of it (:: v negative).

Effect of medium on the focal length and image distance for spherical mirror:

The focal length of a spherical mirror is fixed, i.e., independent of the surrounding media. This is because the law of reflection is invariant even if the medium changes.

For a spherical mirror, if object distance u, image distance v, and focal length f, then the relation between them is  \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)= \(\frac{2}{r}\).

For a fixed object distance u, the image distance v is fixed, because f is fixed and independent of medium. So, if the position of the object and mirror are kept fixed and the surrounding medium is changed no change position of the image occur.

Conjugate Foci or Conjugate Points We have the mirror equation:

We have the mirror equation

⇒ \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)= \(\frac{2}{r}\).

If u and v are interchanged, the equation remains the same. This implies that if the object is placed at the position of the image, the image will be formed at the position of the object.

These two points are called conjugate foci and the above equation is alternatively called the conjugate foci relation.

In the case of the virtual image, conjugate foci are situated on two opposite sides of the mirror, and in the case of a real image, conjugate foci are situated on the same side of the mirror.

Newton’s Equation:

Relation among u, v, f with reference to the pole is

⇒ \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)= \(\frac{2}{r}\).

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

Or, uv-uf-vf+ = 0

Or, uv-uf-vf+f² = 0

Or, u(v-f)-f(v-f) = f²

Or, (u-f)(v-f) = ƒ²

Now, if the object distance and the image distance are measured from the focus, and taken equal to x and y respectively, then we can write,

u-f = x and v-f = y

So, from equation (1) we get,

xy = f²

This equation is known as Newton’s equation. Since ƒ is constant, the graph of x versus y will be a rectangular hyper-bola

Since, f² is a positive quantity, x and y must have the same sign, i.e., the object and the image must be on the same side of the focus.

Magnification of an Image Formed by Spherical Mirrors

Linear or lateral magnification Definition:

The ratio of the height of the image to the height of the object measured in planes that are perpendicular to the principal axis is called the linear or lateral magnification of Linear the image

Linear or lateral magnification is denoted by m

The ratio of the height of the image to the height changed, with no change in the position of the image.

∴ m = \(\frac{\text { height of the image }}{\text { height of the object }}=\frac{I}{O}\) …………………….(1)

1. In the case of a real image formed by a concave mirror:

The ray diagram for the formation of a real image by a concave mirror is shown. Here object distance, RQ-u; image distance, Rq=v, height of the object, PQ = O and height of the image, pq = -I. The Δ PQR and ΔpqR are similar.

∴ \(\frac{p q}{P Q}=\frac{q R}{Q R}\)

Or, \(\frac{-I}{O}=\frac{-v}{-u}\)

Or, \(\frac{I}{O}=-\frac{v}{u}\) ………………(2)

2. In case of a virtual image formed by a concave mirror:

The ray diagram for the formation of a virtual image by a concave mirror is shown.

Here object distance, PQ = -u; image distance, Rq = v, height of the object, PQ =  O and height of the image, pq= I. From ΔPQR and ΔpqR are similar.

∴ \(\frac{p q}{P Q}=\frac{q R}{Q R}\)

Or, \(\frac{I}{O}=\frac{v}{-u}\)

Or, \(\frac{I}{O}=-\frac{v}{u}\) ………………(3)

3. In the case of a virtual image formed by a convex mirror:

In Again the equation can be written as, the ray diagram for the formation of a virtual image by a convex mirror is shown. Here object distance, RQ-u; image distance Rq= y; height of the object, since, PQ = O and height of the image pq = I.

According to the PQR and pqR are similar

∴ \(\frac{p q}{P Q}=\frac{q R}{Q R}\)

Or, \(\frac{I}{O}=\frac{v}{-u}\)

Or, \(\frac{I}{O}=-\frac{v}{u}\) ………………(4)

From equations (2), (3), and (4) it follows that the magnification produced by both kinds of mirror is given by

m= \(\frac{I}{O}=-\frac{v}{u}\) ………………(5)

Some useful hints:

1. The relation m.= – \(\frac{v}{u}\)– is applicable both for concave and convex mirrors.
2. In solving numerical problems, values of u, v, and f should be put with appropriate signs in the mirror equation. No sign for the unknown quantity should be used.
3. Using the appropriate sign of u and v, if
   • m becomes negative, the image will be inverted
   • m becomes positive, the image will be erect
4.  If |m|>1; the size of the image> size of the object
   • If |m|<1; the size of the image< size of the object
   • If |m| = 1; the size of the image size of the object

Magnification in terms of focal length, object distance, and image distance:

From the mirror equation, we get,

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

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

Or, \(\frac{u}{v}\) = \(\frac{u-f}{f}\)

Or, \(\frac{u}{v}\) = \(\frac{f}{u-f}\)

Since, m= – \(\frac{u}{v}\)

∴ m = ( \(\frac{f}{u-f}\) )

Or, ( \(\frac{f}{f-u}\) )

Again the equation can be written as,

1+ \(\frac{v}{u}\)  = \(\frac{v}{f}\)

Or, \(\frac{v}{u}\) = \(\frac{v-f}{f}\)

m= – \(\frac{v-f}{f}\) = \(\frac{f-v}{f}\)

Areal Magnification Definition

Areal magnification for spherical mirrors is the ratio between the image of an area of a plane, and the area of that plane placed perpendicular to the principal axis of the mirror.

Let us consider, that the length and breadth of a plane of a rectangular object are 1 and b respectively.

∴ Area of the plane, A = lb

If the linear magnification of the image of the object by spherical mirror be m, then

length of the image, l’ = m × 1

and breadth, b’ = m × b

∴ Area of the image, A’ = l’b’ = m²lb = m²A Therefore, areal magnification,

m’ = \(\frac{A^{\prime}}{A}\) = m²

Longitudinal or axial magnification of the image of an object kept along the principal axis Definition:

If any object is placed along the principal axis of a spherical mirror then the ratio of the image length and object length is the longitudinal or axial magnification of that image.

Let an object ADEB is placed in front of a spherical mirror MM’.

From the figure, the distance of farther point A of an object, OA = u₁ and that of the nearer point B of the object, OB = u₂

Now, the distance of the farther point of the image, OB’ =v₁ that of the nearer point, OA’=v₂

Longitudinal Magnification , m= \(\frac{v_1-v_2}{u_1-u_2}=\frac{\Delta v}{\Delta u}\)

Here, Δu and Δv are the lengths of the object and its image respectively, along the principal axis of the mirror. For very small magnitudes of Au and Av, these can be considered as du and du respectively.

So,  m” = \(\frac{d v}{d u}\)

Differentiating the equation \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

\(-\frac{1}{v^2} \frac{d v}{d u}-\frac{1}{u^2}\) = 0

Where, [f is constant]

Or, \(\frac{d v}{d u}=-\frac{v^2}{u^2}\)

∴ m” = \(\frac{d v}{d u}\) = – m²

∴  Longitudinal magnification = -(linear magnification)²

Note that, if object length and image length are very small dv then,

∴ m” = \(\frac{d v}{d u}\)

Im may be positive or negative but m” is always negative. This implies that irrespective of whether the object is virtual or real, the image is formed along the principal axis, with opposite alignment. This pianomeonon is called axial inversion.

Formation of Image of a Virtual Object: 

 

In a beam of converging rays is incident on the mirror. In the absence of the mirror, in either case, the

A converging beam of rays would meet at P behind the mirror. But the beam of rays meets at P’ after reflection. Here point P is the virtual object and P’ is the real image of point P. Obviously, object distance OP is positive.

Thus, a convex mirror can also form a real image, but only if the object is virtual. Now consider the case, where the virtual object distance OP is greater than the focal length OF of a convex mirror.

In this case, image P’ will be virtual. with respect to u, we get, In the case of the concave mirror, a real image always be formed for a virtual object and this image is situated between the pole and the focus of the mirror.

Reflection Of Light – Method Of Identifying Mirrors

If an object is placed in front of a plane mirror, a virtual, erect image of the same size as the object is formed. If an object is placed very close to a concave mirror a virtual, erect, and magnified image is formed. A convex mirror forms a virtual, erect image smaller than the object.

For identification, one can hold a pen or a finger very close to the mirror. If an erect image of the same size as the object is formed, then the mirror is a plane one. If an erect image larger than the object is formed, the mirror is concave. If it is smaller than the object the mirror is convex.

Reflection Of Light – Spherical Aberration And Its Remedy

Spherical mirror:

Spherical aberration: The mirror equation is applicable only for spherical mirrors of small aperture to increase the intensity of illumination of the reflected rays to get a brighter image, we often use spherical mirrors of large aperture.

If a beam of rays is incident parallel to the principal axis of a concave mirror of large aperture, not all the rays meet after reflection at a single point. Rather, the reflected rays meet at various points of the principal axis between F to F₁.

So the image becomes indistinct. The larger the aperture of the mirror, the more indistinct the image. This defect of the image is called spherical aberration.

Remedy for spherical aberration: 

If the shape of the mirror is changed from spherical to parab- poloidal, it is possible to get an image free from spherical aberration. Because, according to the geometrical properties of the parabola, all the rays parallel to the principal axis of a paraboloidal mirror meet at its focus after reflection from it.

Reflection Of Light  – Uses Of Spherical Mirrors

Concave mirror:

• Concave mirrors are often used as shaving glasses (mirrors) to see the magnified image of the face, its distance being less than the focal length of the mirrors.
• Small concave mirrors are used by doctors to focus a parallel beam of light on the affected parts like the eye, ear, throat etc. to examine them.
• A small electric lamp placed at the focus of a concave mirror produces a parallel beam of light. So concave mirrors are used as reflectors in torches and car headlights. For better results, paraboloidal mirrors can be used as car headlights. Concave mirrors are used in solar cookers.
• The Palomar Observatory in California has the best reflecting telescope which uses a concave mirror for studying distant stars.

Convex mirror:

Convex mirrors are used as rear-view mirrors in automobiles and other vehicles, designed to allow the driver to see through the rear windshield. This is because a convex mirror forms erect and diminished images of objects and give a wider field of view compared to that of a plane mirror.

Paraboloidal mirror:

According to the geometrical properties of the parabola, the rays parallel to the principal axis of a paraboloidal mirror meet at its focus after reflection from the mirror.

So, if a source of light is placed at the focus F of a paraboloidal mirror, the reflected rays proceed ahead parallel to the principal axis.  For this reason, paraboloidal mirrors are used in car headlights and searchlights.

Sign rules that have been followed here:

To solve the numerical problems, a few sign conventions have been followed here.

• The value of u, v, f, or r  is used with their proper sign-in in the mirror equation, viz \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}=\frac{2}{r}\)
• The focal length for the concave mirror is considered negative and that for the convex mirror is positive.
•  If the image distance is negative, it indicates that the image is formed on the same side as the object, and the image is real and inverted.
• If the image distance is positive, it indicates that the image is formed behind the mirror, and the image is virtual and erect.
• Though mis the standard formula for any type of spherical mirror, to find u or v from this equation, we only take the mod value of m.
• For the determination of the nature of the image, to find the value of m, the associated formula is used following the proper sign convention of u and v. If m becomes positive, the image will be erect, and m becomes negative, the image will be inverted.

Reflection Of Light Numerical Examples

Example 1.  An object is placed 60 cm away from a convex mirror. The size of the image is rd the size of the object. Determine the radius of curvature of the mirror.
Solution:

We have from the mirror equation,

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

According to the question

m = \(\left|-\frac{v}{u}\right|=+\frac{1}{3}\)

v = \(\left|-\frac{u}{3}\right|=\left|\frac{60}{3}\right|\)

U = 60 cm

v= 20 cm

Now, substituting the values of u and v with their proper sign in equation (1) we get,

f = \(\frac{-60 \times(+20)}{-60+20}\)

= + 30 cm

The radius of curvature of the spherical mirror,

r = 2f = 2 × (+30) = +60 cm cm

Example 2. An object of size 5 cm is placed on the principal axis of a convex mirror at a distance of 10 cm from it. The focal length of the convex mirror is 20 cm. Determine the nature, position, and size of the image formed.
Solution:

u=-10 cm; f = +20 cm

The focal length of the convex mirror is positive

Mirror equation:

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

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

= \(\frac{1}{20}+\frac{1}{10}\)

= \(\frac{1+2}{20}\)

= \(\frac{3}{20}\)

Or, v=\(\frac{20}{3}\)

= 6.67 cm

From the positive sign, it can be inferred that the image is formed 6.67 cm behind the mirror. So the image is virtual.

Magnification, m = – \(\frac{v}{u}=-\frac{20 / 3}{-10}\)

= \(\frac{2}{3}\)

As magnification is positive, the image is erect.

The size of the image = \(\frac{2}{3}\)  × 5

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

= 3.33 cm.

Example 3. The image of an object placed 50 cm in front of a concave mirror is formed 2 m behind the mirror. Deter- mine its principal focus and radius of curvature.
Solution:

Here, u=-50 cm; v = +2 m = +200 cm [since the image is formed behind the mirror, v is positive]

We have,  \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

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

= \(\frac{-50 \times(+200)}{-50+(+200)}\)

= \(\frac{-200}{3}\)

= – 66. 67 cm

∴ r = 2f

= 2 × \(\frac{-200}{3}\)

= – \(\frac{-400}{3}\)

= – 133.3 cm

So, the focal length of the concave mirror = 66.67cm and its radius of curvature = 135.3 cm

Example 4.  An object of length 5 cm is placed perpendicularly on the principal axis at a distance of 75 cm from a concave mirror. If the radius of curvature of the mirror is 60 cm, calculate the image distance and its height.
Solution:

Here, u=-75 cm; r = -60 cm;

∴ f =  \(\frac{r}{2}\)

= \(\frac{-60}{2}\)

=-30 cm

We know,    \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}+\frac{1}{-75}=\frac{1}{30}\)

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

= – \(\frac{1}{30}+\frac{1}{75}\)

= – \(\frac{1}{50}\)

Or, v= -50 cm

So, the image is formed at a distance of 50 cm in front of the
mirror. The image is real.

Again , m= \(\frac{\text { height of the image }(I)}{\text { height of the object }(O)}=\frac{v}{u}\)

here we take only the mod value of m as we no need to know the natire of image thus formed

Or, \(\frac{I}{5}=\frac{50}{75}\)

Or, I = 5 × \(\frac{2}{3}\)

= 3.33 cm

So, the height of the images = 3.33cm.

Example 5. A beam of converging rays is incident on a convex mirror of focal length 30 cm. In the absence of the mirror, the converging rays would meet at a distance of 20 cm from the pole of the mirror. If the mirror is situated at the said position where will the converging rays meet? Solution: In the absence of the mirror the converging rays would meet at P. So, P is the virtual object in the case of the mirror.
Solution:

In the absence of the mirror, the converging rays would meet at P. So, P is the virtual object in case of the mirror

In the case of the virtual object, OP = u = +20 cm , f = +30 cm, image distance, v =?

We know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}+\frac{1}{20}=\frac{1}{30}\),

Or, \(\frac{1}{v}\) = \(-\frac{1}{20}+\frac{1}{30}\),

= \(\frac{-1}{60}\)

Or, v= -60 cm

As v is negative, the converging rays will meet at Q at a distance
of 60 cm in front of the mirror

Example 6. An image of size \(\frac{1}{n}\) times that of the object is formed in 1 P a convex mirror. If r is the radius of curvature of the ‘HIJL 1 mirror, calculate the object distance.
Solution:

Considering the mod value only, magnification,

m = \(\frac{1}{n}\)

= \(\frac{v}{u}\)

Or, v = \(\frac{u}{n}\)

We know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\)

[ Here object distance = -u]

Or, \(\frac{n}{u}-\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{(n-1)}{u}=\frac{1}{f}\)

= (n-1)f

Again, f= \(\frac{r}{2}\) , Then u= (n-1) \(\frac{r}{2}\)

∴ Object distance = (n-1)\(\frac{r}{2}\)

Example 7. An object of height 2.5 cm is placed perpendicularly on the principal axis of a concave mirror of focal length f at a distance off. What will be the nature of the image of the object and its height?
Solution:

Here, u = \(\frac{3}{4}\) f= Focal Length = -f

We know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}-\frac{4}{3 f}=-\frac{1}{f}\)

Or, \(\frac{1}{v}=-\frac{1}{f}+\frac{4}{3 f}\)

= \(+\frac{1}{3 f}\)

Or, v= +3f

The positive sign of v indicates that the image formed by the concave mirror is virtual in nature and is situated behind the mirror at a distance of 3f.

Magnification, m = – \(\frac{v}{u}\)

= \(-\frac{3 f}{-\frac{3}{4} f}\)

= 4

Again, m= \(=\frac{\text { height of the image }}{\text { height of the object }}\)

Or, 4 = \(\frac{\text { height of the image }}{2.5}\)

∴ Height of the image = 10 cm

Example 8.  The image of the flame of a candle due to a mirror is formed on a screen at a distance of 9 cm from the candle. The image is magnified 4 times. Determine the nature, position, and focal length of the mirror.
Solution: 

As a magnified image image is formed on a screen, the image is real.

Let u = – x cm and hence v = -(x+9) cm

Here, m = 4 or \(\frac{v}{u}\) = 4 [taking only the mod value of m]

Or, \(\frac{x+9}{x}\)  = 4

Or, 4x = x+9 or, x = 3

∴ u = -3 cm

So, the mirror is situated at a distance of 3 cm from the flame

Now, v= -(3+9) = -12 cm

We, know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\) .

Or, \(\frac{1}{-12}+\frac{1}{-3}=\frac{1}{f}\) .

Or, \(-\left(\frac{1+4}{12}\right)=\frac{1}{f}\) .

or, f = \(-\frac{12}{5}\)

= -2.4 cm

So, the focal length of the mirror is 2.4 cm. Its negative sign indicates the nature of the mirror as a concave one [Fig. 1.40].

Example 9. The focal length of a concave mirror is f. A point object is placed beyond the focal length at a distance from the focus. Prove that the image will be formed beyond the focal length at a distance  \(\frac{f}{x}\)from the focus and the magnification of the image will be (\(\frac{1}{x}\)
Solution:

Here, object distance, u = -(f+xf)

We, know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}+\frac{1}{-(f+x f)}\)

Or, \(\frac{1}{v}=\frac{1}{f+x f}-\frac{1}{f}\)

= – \(\frac{1}{v}=\frac{1}{f}\left[\frac{1}{1+x}-1\right]\)

= \(\frac{-x}{f(1+x)}\)

∴ v =  \(\frac{f(1+x)}{-x}=-\left(f+\frac{f}{x}\right)\)

So, the image is formed beyond the focal length at a distance of be u. \(\frac{f}{x}\) from the focus

∴ Magnification , m= \(-\frac{v}{u}\)

=- \(\frac{f(1+x)}{x}{-f(1+x)}\)

=  \(\frac{1}{x}\)

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

Example 10. A thin glass plate is placed in between a convex mirror of a length 20 cm and a point source. The distance between the glass plate and the mirror is 5 cm.  The image formed by the reflected rays from the front face of the glass plate and that due to the reflected rays by the convex mirror coincides at the same point. What is the distance of the glass plate from the source? Draw the ray diagram

The ray diagram has been shown. The distance between the glass plate M and the point source

P = x cm (say). Light rays starting from P form the image at Q by reflection at the convex mirror. The glass plate also forms the image of P at Q.

∴ Distance of the image from the glass plate = x cm

∴ Distance of the image from the convex mirror = (x-5) cm glass plate also forms the image of P at Q.

∴ Distance of the object from the convex mirror=-(x+5) cm 

We know , \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

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

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

Or, \(\frac{10}{x^2-25}=+\frac{1}{20}\)

∴ The distance of the glass plate from the source = 15 cm

Example 11. The sun subtends an angle 0.5° at the center of a concave mirror having a radius of curvature 1 m. What will be the diameter of the image of the sun formed by the mirror?
Solution:

MM’ is a concave mirror Let the diameter of the sun be D and the distance of the sun from the mirror be u,

∴ \(\frac{D}{u}=\frac{\pi}{360}\)

0.5 ° = \(\frac{1}{2} \times \frac{\pi}{180}=\frac{\pi}{360}\) radian

Or, \(\frac{360 D}{\pi}\)

The sun is at a large distance from the mirror. So, the rays coming from the sun are assumed to be parallel and hence its image will be formed at the focal plane.

∴ v= \(f\frac{r}{2}\)

= \(\frac{1}{2}\)

= 0.5 cm

Now magnification, m = \(-\frac{v}{u}=\frac{-0.5 \pi}{-360 D}\)

⇒ \(\frac{0.5 \pi}{360 D}\)  [since in this case both u and v are negative]

Again \(\frac{I}{O}=\frac{\text { diameter of the image of the sun }}{\text { diameter of the sun }(D)}\)

∴  Diameter of the image of the sun

= m × D = \(\frac{0.5 \pi}{360 D}\) × D= 0.004363

m = 0.4363 cm. [ taking only the numerical of m]

Example 12. An object is placed at a distance of 25 cm from a con- cave mirror and a real image is formed by the mirror at a distance of 37.5 cm. What is the focal length of the mirror? Now if the object is moved 15 cm towards the mirror, what will be the image distance, its nature and magnification?
Solution:

Here, u = -25 cm, v = -37.5 cm

We know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{-37.5}+\frac{1}{-25}=\frac{1}{f}\)

= \(-\frac{10}{375}-\frac{1}{25}\)

= \(-\frac{25}{375}=-\frac{1}{15}\)

Or, f= – 15cm

Now if the object is moved 15 cm towards the mirror, the object distance will be (25-15) = -10 cm

=  \(\frac{1}{v}+\frac{1}{-10}=-\frac{1}{15}\)

Or, \(-\frac{1}{15}+\frac{1}{10}\)

= \(\frac{-2+3}{30}=+\frac{1}{30}\)

= + \(\frac{1}{30}\)

Or, v= + 30 cm

The positive sign of v indicates that the image formed by the mirror is virtual and it is formed at a distance of 30 cm behind the mirror.

Magnification, m = \(-\frac{v}{u}\)

= \(-\frac{30}{-10}\)

= 3

∴ The image will be magnified 3 times. The positive sign of m indicates the image as erect.

Example 13. An object is placed at a distance of 50 cm in front of a convex mirror. Now a plane mirror is placed in between the object and the convex mirror, covering the lower half of the convex mirror. If the distance of the plane mirror from the object is 30 cm, it is seen that there is no parallax of the images formed by the two mirrors. What is the radius of curvature of the convex mirror?
Solution:

MM’ is a convex mirror and A is a plane mirror. P’Q’ is the object and P’Q’ is the image formed at the same place by the two mirrors.

The radius of curvature,

Here OR = 50 cm , AR = 30 cm

Image distance from the plane mirror = O’A = 30 cm

According to the figure,

OA = OR-AR

= 50-30

= 20 cm

∴ Image distance from the convex mirror,

O’O = O’A-OA = 30-20 = 10 cm

We have \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Here, v= +10 cm , u= -50 cm

∴ \(\frac{1}{10}+\frac{1}{-50}=\frac{1}{f}\)

Or,  \(\frac{4}{50}=\frac{1}{f}\)

Or,  \(f=+\frac{25}{2}\)

Radius of curvature,

r = 2f

\(\frac{25}{2}\)

=  25 cm

The radius of curvature of the convex mirror = 25 cm

Example 14. A concave mirror of focal length 10 cm and a convex mirror of focal length 15 cm are held co-axially face to face at a distance 40 cm apart. An object of height 2 cm is placed perpendicularly on the common axis in between the two mirrors. The distance of the object from the concave mirror is 15 cm. Considering the first reflection occurs in the concave mirror and the second reflection in the convex mirror, calculate the position, nature, and height of the final image.
Solution:

In the case of the first reflection in the concave mirror: u = -15 cm; f = -10 cm

We know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{y_1}+\frac{1}{-15}=\frac{1}{-10}\)

Or, \(\frac{1}{v}=-\frac{1}{10}+\frac{1}{15}\)

Or, \(\frac{3_1+2}{30}\)

Or, \(-\frac{1}{30}\)

v= -30 cm

So, the image formed by the concave mirror is real and formed at a distance of 30 cm from the mirror. This image acts as the object of the convex mirror.

In case of the second reflection in the convex mirror:

u = -(40- 30) = -10 cm , f = +15 cm

We know \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}+\frac{1}{-10}=+\frac{1}{15}\)

Or, \(\frac{1}{v}=+\frac{1}{15}+\frac{1}{10}\)

= + \(\frac{1}{6}\)

Or, v= + 6cm

So the final image is virtual and is formed at a distance of 6 cm
behind the convex mirror.

Magnification by the concave mirror,

m₁ = \(\frac{v}{u}\)

= \(\frac{-30}{-15}\)

= -2

Magnification by the convex mirror,

m₂ = – \(\frac{v}{u}\)

= – \(\frac{6}{-10}\)

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

∴ Total magnification.

m = m₁ × m₂

= -2 ×\(\frac{3}{5}\)

=  – \(\frac{6}{5}\)

As m is negative, so the final image is inverted with respect to the object.

∴ Height- of the final image (taking only the magnitude of m)

= \(\frac{6}{5}\) × height of the object

= \(\frac{6}{5}\) × 2

= \(\frac{12}{5}\)

= 2. 4 cm

Example 15. The focal length of a concave mirror is 30 cm. An object is placed at a distance of 45 cm in front of the mirror. A plane mirror is placed perpendicularly on the principal axis of the concave mirror in such a way that the object is situated in between the two mirrors. Light rays from the object at first are reflected from the concave mirror and then from the plane mirror. As a result the final image coincides with the object. What is the distance between the two mirrors?
Solution:

In case of the concave mirror M₁ M₂, P is the object and P’ is its image.

Here, u = -45 cm,  f= – 30 cm

We know \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}-\frac{1}{45}\)

= \(-\frac{1}{30}\)

Or, \(\frac{1}{v}=-\frac{1}{30}+\frac{1}{45}\)

= \(\frac{-3+2}{90}\)

= \(\frac{1}{90}\)

Or, v= -90 cm

Op’ = 90 cm

According to the question, the reflected rays from the concave mirror form the final image at P after reflection from the plane
mirror M.

∴ PM=  P’M = x (say)

Here, OP’ = OP+PM+P’M

Or, 90 = 45+x+x

Or, x = \(\frac{45}{2}\)

= 22.5 cm

∴ Distance between the mirrors = 45+22.5 67.5 cm.

Example 16. A concave mirror forms a real image magnified two times. If both the object and the screen are moved a real Image magnified three times that of the object is formed. If the screen is moved through a distance of 25cm, then determine the displacement of the object and focal length of the mirror.
Solution:

In the first case,

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

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

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

∴ 1+m = –  \(-\frac{v}{u}\)…………………….(1)

Since m= – \(-\frac{v}{f}\)

In the first case, magnification = 2

∴ 1+2 = \(-\frac{v}{f}\)

Or,  – \(\frac{v}{f}\) = 3…………………….(2)

In the second case, magnification = 3

∴ 1+3 = – \(\frac{(v+25)}{f}\)

Or,  4 = – \(\frac{v}{f}-\frac{25}{f}\)

∴ 4= 3 – \(3\frac{25}{f}\)

Or, f= – 25 cm

The focal length of the concave mirror = 25cm

From equation (2) we have,

v =3f = -3 × (-25) = 75 cm

According to the equation of a spherical mirror,

⇒ \(\frac{1}{u}=\frac{1}{f}-\frac{1}{v}=\frac{1}{75}-\frac{1}{25}\)

[since the image is real, hence v is taken negative]

Or,

u = -37.5 cm

In the first case object distance 37.5 cm

In the second case image distance

v₁= v+25= 75+25= 100 cm

Suppose, object distance = u₁

Since in this case magnification =3

∴ \(u_1=-\frac{1}{3} v_1\)

= \(\frac{100}{3}\)

= – 33.33 cm

∴ Object distance in the second case =33.33cm

∴ Displacement of the object = 37.5-33.33

= 4.17cm

So, the displacement of the object = 4.17 cm, and the focal length of the mirror = 25 cm

Example 17.  A cube of side 2 m is placed in front of a large concave mirror of focal length 1 m in such a way that the face A of the cube is at a distance of 3 m and the face B at a distance of 5 m from the mirror.

1. Calculate the distance between the images of the faces A and B.
2. Determine the height of the images of faces A and B. 
3. Will the image of the cube be a cube

Solution:

1. Distance of the face A from the mirror, u₁ = -3 m, focal length of the mirror, f= -1 m

Let the image distance of the face A from the mirror be v₁ m

We know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or,   \(\frac{1}{v_1}-\frac{1}{3}=-\frac{1}{1}\)

Or,= v₁ = -1.5 m

Distance of the face B from the mirror, u=-5 m; the image distance of this face = v₂ m

We know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v_2}-\frac{1}{5}=-\frac{1}{1}\)

Or,= v₂ = -1.25 m

So, the distance between the images of the faces A and B

= v₁– v₂ = 1.5-1.25 = 0.25 m [Taking magnitude of v₁ and v₂ ]

2. Magnification in the first case

m₁ = \(\frac{1}{v_1}-\frac{1}{u_1}\)

= \(\frac{-1.5}{-3}\)

= – 0.5

∴ \(\frac{\text { height of the image of the face } A\left(I_A\right)}{\text { length of the face } A\left(O_A\right)}\)

Or, I_A= -5.5 × 2 = -1m

Again, magnification in the second case

⇒ \(\frac{1}{v_2}-\frac{1}{u_2}\)

m₂ = \(\frac{1}{v_2}-\frac{1}{u_2}\)

= \(\frac{-1.25}{-5}\)

= – 0.25

∴ Height of the image of the face B,

Or, I_B= -5.5 × 2 = -1m

I_B =  m₂ ×  height of the face B

= -0.25 × 2

= -0.5 m

3. So, it is seen that the height of the image of face A and that of the face B are not equal. So, the image of the cube will no longer be a cube.

Example 18. A is a point object in a circular track. A light ray starting from object A is reflected twice by the circular track and returns again to A. The angle of incidance is a. The distance of A from the center of the circular track is x and the diameter of the circular track from A intersects the path of the ray at a point D whose distance from the center of the circular track is y.
Show that, tan α = \(\sqrt{\frac{x-y}{x+y}}\)

Solution:

In Δ OBC, OB = OC; so, ∠OBC= ∠OCB = α

Also ∠ABC = ∠ACB= 2α; so, AB = AC

Since Δ ABC is isosceles, hence median AD ⊥ BC

∴ tan α =  \(\frac{y}{B D}\) and tan 2α = \(\frac{x+y}{B D}\)

Or, \(\frac{\tan 2 \alpha}{\tan \alpha}=\frac{x+y}{y}\)

Or, \(\frac{2 \tan \alpha}{\tan \alpha\left(1-\tan ^2 \alpha\right)}=\frac{x+y}{y}\)

Or, \(1-\tan ^2 \alpha=\frac{2 y}{x+y}\)

Or, \(\tan ^2 \alpha=1-\frac{2 y}{x+y}\)

= \(\frac{x-y}{x+y}\)

Or, tanα = \(\sqrt{\frac{x-y}{x+y}}\)

So, the distance between the images of the faces A and B

Example 19. A concave mirror and a convex mirror are placed co-axially face to face. The focal length of each of them is f and the distance between them is 4f. A point source is so placed on their common axis in between the two mirrors that if the first reflection is considered to take place on the convex mirror, the final image coincides with the point source. Determine the position of the source.
Solution:

Let the point source O be situated at a distance x from the convex mirror

If the image distance is v₁ then

⇒ \(\frac{1}{v_1}+\frac{1}{-x}=\frac{1}{f}\)

Or, \(\frac{1}{v_1}=+\frac{1}{f}+\frac{1}{x}\)

= \(+\frac{1}{f}+\frac{1}{x}=\frac{(x+f)}{f x}\)

Or, \(v_1=\frac{f x}{x+f}\)

This image will be formed at Oj behind the convex mirror. Now u X this image will act as the object of the concave mirror.

∴ Object distance for the concave mirror,

u₂= 4f+ v₁

= \(4 f+\frac{f x}{x+f}\)

= \(\frac{5 f x+4 f^2}{x+f}\)

∴ Image distance v₂ = 4f- x

Since the final image coincides with the object

∴ \(\frac{1}{v_2}+\frac{1}{u_2}=\frac{1}{f}\)

Or, \(\frac{1}{4 f-x}+\frac{x+f}{5 f x+4 f^2}=\frac{1}{f}\)

Or, \(\frac{1}{4 f-x}=\frac{1}{f}-\frac{x+f}{5 f x+4 f^2}\)

= \(\frac{4 x+3 f}{f(5 x+4 f)}\)

Or = x² – 2fx – 2f² = 0

Or, \(\frac{2 f \pm \sqrt{4 f^2+8 f^2}}{2}=f \pm f \sqrt{3}\)

∴ x = \(f(1+\sqrt{3})\)  [neglecting the negative value]

The distance of the source from convex mirror = \(f(1+\sqrt{3})\)

Example 20.  The diameter of the moon is 3450 km and its distance from the earth is 3.8 x 105 km, What will be the diameter of the image of the moon formed by a convex mirror of focal length 7.6 m? As the moon is very far away from the earth the image of the moon will be formed in the focal plane of the mirror
Solution:

Hence, v = f = 7.6 m

Now, u= 3.8 × 105 km and magnification, m = \(\frac{v}{u}\)

[Taking the mod value of m]

∴ The diameter of the image

= m × diameter of the moon

= \(=\frac{7.6 \mathrm{~m}}{3.8 \times 10^5 \mathrm{~km}} \times 3450\)km

= 00.69m

= 6.9 cm

Example 21. Three times magnified image of an object is formed on a screen placed at a distance of 8 cm from the object with the help of the spherical mirror. Determine the nature of the mirror, focal length, and the distance of the mirror from the object.]
Solution:

Since the image is formed on a screen, the image is real and the mirror is concave.

Suppose, u= x cm and so v = (x+8)cm

Here,m = 3 or, \(\frac{v}{u}\) = 3 [taking the mod value of m]

Or, \(\frac{v}{u}\) = 3

Or, \(\frac{(x+8)}{x}\) = 3

Or,   x= 4cm

So the mirror is situated at a distance of 4 cm from the object

∴ v = (4+8) = 12 cm

We know , \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or,  \(\frac{1}{-12}+\frac{1}{-4}=\frac{1}{f}\)

Or, f= -3 cm

∴  Focal length of the mirror

= 3 cm. A negative sign in the value of/ supports the mirror as concave in nature

Example 22. An object is placed just at the middle point between a concave mirror of radius of curvature 40 cm and a convex mirror of radius of curvature 30 cm facing each other. The mirrors are situated 50 cm apart from each other. Considering the first reflection occurs in the concave mirror, determine the position and nature of the image formed by this mirror. Next, find the position and nature of the image formed by the convex mirror considering the first image of concave mirror as its object.
Solution:

Focal length of the concave mirror = \(\frac{40}{2}\) = 20 CM

Focal length of the convex mirror = \(\frac{30}{2}\)  = 15cm

In case of the concave mirror, u= – 25 cm: f= -20 cm.

We know, \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}+\frac{1}{-25}=\frac{1}{-20}\)

Or, \(\frac{1}{v}=\frac{1}{25}-\frac{1}{20}=\frac{4-5}{100}\)

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

Or, v= – 100 cm

So, the image formed by the concave mirror in the

absence of the convex mirror would be real and inverted and situated at Q₁ at a distance of 100 cm from the concave mirror. This image will act as a virtual object for the convex mirror.

Her O₁Q₁ = 100 cm

Now for the presence of the convex mirror M₃M₄ object distance, M₃M₄ = , object distance , u₁= (100-50) cm = 50 cm;

Focal length, f₁= 15 cm.

According to the equation of mirror,

⇒ \(\frac{1}{v_1}+\frac{1}{50}=\frac{1}{15}\)

Or, \(\frac{1}{v_1}=\frac{1}{15}-\frac{1}{50}\)

Or, \(\frac{10-3}{150}=\frac{7}{150}\)

Or,  v₁= \(\frac{150}{7}\)

= 21.43cm

So, the final image will be formed at Q₂ at a distance of 21.43 cm from the convex mirror. A positive value of v₁ indicates that this image is virtual.

Example 23. An arrow of height 2.5 cm is situated vertically at a distance of 10 cm from a convex mirror of a focal length 20 cm. Where will the image be formed? Determine its height. If the arrow is moved away from the mirror what will happen to its image?
Solution:

Here, u = -10 cm and f= 20 cm

We know,  \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}+\frac{1}{-10}=\frac{1}{20}\)

Or, \(\frac{1}{v}=\frac{1}{10}+\frac{1}{20}\)

Or, \(\frac{2+1}{20}=\frac{3}{20}\)

or, v= \(\frac{20}{3}\)

= 6. 66cm

So, the image will be formed at a distance of 6.66 cm behind the convex mirror.

Magnification, m =\(\frac{I}{O}=-\frac{v}{u}\)

I = \(-O \times \frac{v}{u}\)

= \(-2.5 \times \frac{20}{3 \times(-10)}\)

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

= 1.66cm

So, the height of the image = 1.66 cm. If the arrow is moved away from the convex mirror the image will move towards the focus of the mirror and will be diminished in size

Example 24. A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. What is the length of the image?
Solution:

Here for the side A, f = -10 cm, u = -20 cm.

From the equation, \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\) we get v = -20 cm

For the side B, u – -30 cm

Similarly, v = -15 cm

Length of the image = 20- 15 = 5 cm

Reflection Of Light Synopsis

Reflection of light:

1. Coming through a medium when light incidents on the upper surface of another medium, then a part of that incident light returns to the first medium by changing its direction.
2. This phenomenon is known as the reflection of light.

Laws of reflection:

1. The incident ray, the reflected ray and the normal to the reflecting surface at the point of incidence lie in the same plane.
2. The angle of incidence is equal to the angle of reflection.

Image:

When rays of light diverging from a point source after reflection or refraction converge to or appear to diverge from a second point, the second point is called the image of the first point.

Focus in case of reflection from a spherical surface:

When a beam of rays parallels to the principal axis is incident on a spherical mirror, it is seen that the reflected rays either converge to or appear to diverge from a fixed point on the principal axis. This point is called the principal focus or focal point or briefly the focus of the mirror.

Conjugate foci:

1. If a pair of points is such that when an object is placed at one of them, its image is formed at the other by a fixed mirror, then that pair of points are called conjugate foci of that mirror.

2. The focal length of a concave mirror is taken as positive and the focal length of a convex mirror is taken as negative.

3. The distance measured from the pole of the mirror and in ‘front of it is taken as positive but the distance measured in the back side of the mirror is taken as negative.

4. If the aperture is small then the focal length of a spherical mirror becomes half of its radius of curvature

5. Real image is formed in front of the mirror but virtual is formed at the back of the mirror.

6. The ratio of the length of the image to that of the object is called linear magnification of the image.

7. The ratio of the area of an image to that of the two-dimensional object is called areal magnification.

8. The ratio of the length of the image to that of the object along the principal axis is called longitudinal or axial mag- notification.. da

9. In the case of a spherical mirror longitudinal magnification is equal to the square of the linear magnification.

10. Angle of reflection (r) = angle of incidence (1)

11. Angle of deviation for a ray of light after reflection from a plane mirror,

δ = 180°-2i [i= angle of incidence]

12. In case of the image formed by a plane mirror, the distance of the image from the mirror

= distance of the object from the mirror.

13. In the case of a spherical mirror of a small aperture the relation between the focal length of the mirror () and its radius of curvature (r) is

f = \(\frac{r}{2}\)

14. In the case of a spherical mirror of a small aperture, the relation among the object distance (u), image distance (v), and focal length (f) is

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

15. Newton’s equation:

xy = f² [where, x = u-f and y = v-f]

16. Linear magnification:

m = \(\frac{\text { length or height of the image }(I)}{\text { length or height of the object }(O)}\)

= \(\frac{\text { image distance }(v)}{\text { object distance }(u)}\)

= \(\frac{f}{f-u}=\frac{f-\nu}{f}\)

17. Areal magnification for a two-dimensional object:

m’ = \(\frac{\text { area of image }}{\text { area of object }}\)

18. Longitudinal or axial magnification:

m” = \(\frac{\text { length of the image along the principal axis }}{\text { length of the object along the principal axis }}\)

Rules for solving numerical problems relating to spherical mirrors

To solve the numerical problems in connection with spherical mirrors the following rules are to be followed:

• In the general equation of mirrors, put the numerical values of u, v, f etc. with their proper signs.
• Then after solving the equation, draw inference about the distance from the sign.
• A positive sign of the focal length will suggest that the mirror is convex while a negative sign will suggest that the mirror is concave.
•  If the image distance is negative, we have to understand that the image has been formed in front of the mirror and it is real and inverted.
• Again, if the image distance is positive, we have to understand that the image has been formed behind the mirror and it is virtual and erect.

Reflection Of Light Very Short Question And Answers

Question 1. Can an image be formed due to diffused reflection? 
Answer: No

Question 2. Can a plane mirror form a real image?
Answer: Yes

Question 3. If a light ray is incident on a horizontal plane mirror making an angle 30° with the mirror, what angle will the reflected ray make with the mirror?
Answer: 30

Question 4. What type of reflection takes place on the screen of a cinema hall?
Answer: Diffuse

Question 5. What is the angle of deviation of a ray incident perpendicularly on a reflecting surface?
Answer:  180°

Question 6. Focus of which spherical mirror is virtual.
Answer:  Convex

Question 7. Two concave mirrors have same focal length but different apertures. Both the mirrors form an image of the sun on a screen. For which image formation will the temperature of the screen become higher?
Answer:  Concave

Question 8. The secondary focus of a concave mirror is a fixed point. Is this statement correct?
Answer: No

Question 9. The principal focus of a concave mirror is a fixed point. Is this statement correct?
Answer:  Yes

Question 10. Two concave mirrors have the same aperture but different focal lengths. Both form images of the sun on a screen. For which image formation will the temperature of the
Answer:  In both cases, hotness will be the same

Question 11. The radius of curvature of a concave mirror is 30 cm. What is its focal length?
Answer: 15 cm

Question 12. Which spherical mirror is called a divergent mirror-concave or convex?
Answer:  Convex

Question 13. The focal length of a spherical mirror is 40 cm. What is its radius of curvature?
Answer: 80 cm

Question 14. What is the value of the focal length of a plane mirror?
Answer: Infinite

15. What is the power of a plane mirror?
Answer: Zero

Question 16. What is the relation between the radius of curvature (r) and focal length (f) of a spherical mirror?
Answer: r = 2f

Question 17. Does the size of mirror affect the nature of the image?
Answer: No

Question 18. The focal length of a concave mirror is equal for all colors of light. Is the statement true or false? [true] [yes] 21. What type of mirror is to be used for getting parallel rays from a small source of light?
Answer: True

Question 19. What will happen to the focal length of the concave mirror when it is immersed in water? Will remain unchanged
Answer: Will remain unchanged

Question 20. Where should an object be placed in front of a concave mirror to get a magnified image?
Answer:  Between f and 2f 

Question 21. Why do we sometimes use a concave mirror instead of a plane mirror as a common mirror? [to get a magnified erect image]
Answer: To get a magnified erect image

Question 22. What is the magnification of the image of an object placed at the center of curvature of a concave mirror?
Answer: 1

Question 23. What is the magnification of an object placed at the focus Answer: Infinity

Question 24. Can a convex mirror ever form a real image of a real object? of a concave mirror?
Answer:  No

Question 25. What will the image of an object placed before a convex screen be higher? mirror-erect or inverted?
Answer: Erect

Question 26. Two concave mirrors are placed face to face and they have the same centre of curvature. A point source of light is placed at their common center of curvature. Where will the image be formed?
Answer: At their common center of curvature

Question 27. If an object is placed between the pôle and the focus of a concave mirror, will the size of the image be magnified with respect to the size of the objector?
Answer: Yes

Question 28. What is the minimum distance between an object and its image formed by a concave mirror?
Answer:  Zero

Question 29. If a concave mirror is immersed in water will its focal length change?
Answer: No

Question 30. In the case of a concave mirror, what is the shape of the uv graph?
Answer:  Rectangular hyperbola]

Question 31. In the case of a concave mirror, what is the shape of the graph?
Answer: Straight line

Question 32. The focal length of a concave mirror in vacuum is 2 m. What will be the focal length of the concave mirror in a medium of refractive index 2.76?
Answer: 2 m

Question 33. At what distance in front of a concave mirror (f = 10 m) an object is to be placed so that the size of the image will be halved of the size of the object?
Answer: 30 cm

Question 34. If the conjugate foci of a spherical mirror lie on the same side of the mirror then, what is the nature of the mirror?
Answer: Concave

Question 35. What is the magnification produced in a plane mirror? 40. A cube is placed in front of a large concave mirror. Will the image of the cube be a cube?
Answer: 1

Question 36. Can a convex mirror form a real image?
Answer: Can form real image of a virtual object

Question 37. For a spherical mirror if the linear magnification of the image be m what will be its lateral magnification?
Answer: m²

Question 38. The image of a candle formed by a concave mirror is cast on a screen. What will happen if the mirror is covered partly?
Answer: The brightness of the image will be reduced]

Question 39. Will the focal length of a spherical mirror be affected if the wavelength of the light used is increased?
Answer: No

Question 40. What type of mirror do car drivers use to view the traffic at the back of the car?
Answer: Convex

Question 41. What type of mirror do dentists use?
Answer: Concave

Question 42. What type of mirror is used in searchlights and headlights of vehicles?
Answer: Paraboloidal

Reflection Of Light Fill In The Blanks

Question 1. The reflecting surface from which regular reflection of light takes place is called__________________  object 
Answer:  Smooth plane reflector

Question 2. A plane mirror can form a real image of a____________ object
Answer: Virtual

Question 3. A smaller virtual image is formed by the blank). mirror. [Fill in Answer: Convex

Question 4. An object is moving towards a convex mirror from a large distance. The image will move with _________________ velocity than  the object  ___________  the mirror 
Answer:  Less, towards

Reflection Of Light Assertion Reason Type

Direction: These questions have statement I and statement II. the four choices given below, choose the one that best scribes the two statements.

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

Question 1.

Statement 1: The formula connecting u, v, and ƒ for a spherical mirror is valid only for mirrors whose sizes are very small compared to their radii of curvature.

Statement 2: Laws of reflection are strictly valid for plane surfaces but not for large spherical surfaces.

Answer:  3. Statement 3 is true, statement 2 is false.

Question 2.

Statement 1: A concave mirror is preferred to a plane mirror for shaving.

Statement 2: When a man keeps his face between the pole and the focus of the mirror, an erect and highly magnified virtual image is formed.

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

Question 3. 

Statement 1: A virtual image can not be directly photographed.

Statement 2: A virtual image can be produced by using a convex mirror.

Answer: 2. Statement 2 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 4.

Statement 1: In the absence of diffuse reflection an object would appear either dazzlingly bright or quite dark.

Statement 2: The angle of incidence is not equal to the angle of reflection in this case.

Answer:   3. Statement 3 is true, statement 2 is false.

Question 5.

Statement 1: Convex mirror is used as driver’s mirror.

Statement 2: Convex mirror gives an index wider field of view of the traffic.

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

Reflection Of Light Match The Columns

Question 1. 

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

Question 2. The nature of the image and the type of mirror are given in column A and column B respectively. Match the column.

Answer:  1-D, 2-C, 3 A, 4 – B

Question 3. Some relations related to spherical mirror and their field of application are given in column A and column B respectively (The symbols have their usual meanings). Match the column.

 

Answer:  1- D, 2- C, 3- B, 4- A

Question 4. The position of an object with respect to a concave mirror and the position of the image are given in column 1 and column 2 respectively.

Answer: 1- B, 2-D, 3- E, 4- C, 5-A

Question 5. Match the corresponding entries of column I with column [where m is the magnification produced by the mirror].

1. 1. (A, C ) 2. (A, D)  3. ( A, B)  4. (C, D)
2. 1. (A, D)  2. (B, C) 3. (B, D) 4. (C, D)
3. 1. (C, D) 2. (B, D) 3. (B, C) 4. (A, D)
4. 1. (B, C) 2. (B, C )3. (B, D) 4. (A, D)

Answer: 1. (B, C)  2. (B, C) 3. (B, D) 4. (A, D)

Magnification of image produced by a spherical mirror,

For real images, m<0, and m> 0 for virtual images.

The convex mirror always forms a virtual image, so m> 0 always and the size of the image is less or equal to the size of object 1.e., m≤1.

In the case of the concave mirror, if the real image is formed, then there can be |m|≥ 1 and |m | ≤1, and for a virtual image m>1. The option is correct.