WBCHSE Class 12 Physics Optical Instruments Multiple Choice Questions

Optics

Optical Instruments Multiple Choice Questions

Question 1. A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in

1. A larger angle to be subtended by the object at the eye and hence viewed in greater details
2. The formation of a virtual erect image
3. Increase in the field of view
4. Infinite magnification at the near point

Answer: 1 And 2

Question 2. An astronomical refractive telescope has an objective of focal length 20 m and an eyepiece focal length of 2 cm

1. The length of the telescope tube is 20.02 m
2. The magnification is 1000
3. The image for med inverted
4. An objective of a larger aperture.will increase the brightness and reduce the chromatic aberration of the image

Answer: 1, 2 And 3

Question 3. A normal eye cannot see the object nearby from a distance of 25 cm, as

1. The focal length of the eye is 25 cm
2. The distance between the eye lens and the retina is 25 cm
3. The eye is unable to adjust the distance between the lens and the retina below a certain limit
4. The eye is unable to adjust the focal length of the eye lens below a certain limit

Answer: 4 The eye is unable to adjust the focal length of the eye lens below a certain limit

Read and Learn More Class 12 Physics Multiple Choice Questions

Question 4. The ability of the eye to see objects at all distances is called

1. Binocular vision
2. Myopia
3. Hypermetropia
4. Accommodation

Answer: 4 Accommodation

Question 5. The numerical aperture for a human eye is of the order of

1. 1
2. 0.1
3. 0.01
4. 0.001

Answer: 4 0.001

Question 6. If there had been one eye of the man then

1. The image of the object would have been inverted
2. Visible region would have decreased
3. The image would have not been seen three three-dimensional
4. 2 and 3 both

Answer: 4. 2 and 3 both

Question 7. What type of lens should be used in spectacles for the remedy of myopia?

1. A concave lens whose focal length is equal to the  distance of the far point of the eye defective eye from the lens
2. A convex lens whose focal length is equal to the die distance of the far point of the eye defective eye from the lens
3. A concave lens whose focal length is equal to the distance of the near point of the defective eye from the lens
4. Convex lens whose focal length is equal to the eye distance of the near point of the defective eye from the lens

Answer:  1

Question 8. A person who can see things most clearly at a distance of 10cm, requires spectacles to be able to see clearly things at a distance of 30 cm. What should be die focal length of the spectacles?

1. 15 cm (convex)
2. 15 cm (concave)
3. 10 cm
4. Zero

Answer: 1. 15 cm (convex)

Question 9. The least distance of distinct vision of a man is 45 cm. He uses a lens of focal length 15 cm for reading. The magnification that he gets

1. 4
2. 3
3. 2
4. 1

Answer: 1. 4

Question 10. The image formed by the objective of a compound microscope is

1. Virtual and magnified
2. Virtual and diminished
3. Real and diminished
4. Real and magnified

Answer: 4. Real and magnified

Question 11. To get a large magnification from a compound microscope

1. The focal length of the objective should be large while The focal length of the eyepiece should be small.
2. The focal length of the objective should be small while the focal length of the eyepiece should be large
3. Both of the focal lengths of the objective and the eyepiece should be large
4. Both of the focal lengths of the objective and the eyepiece should be small

Answer:  4. Both of the focal lengths of the objective and the eyepiece should be small

Question 12. The length of the tube of a compound microscope is 21.5 cm. The focal lengths of the objective and the eyepiece are 1.6 cm and 2.1 cm respectively. If the final image is situated at infinity, then the distance of the object from the objective is

1. 3 cm
2. 1.7 cm
3. 6 cm
4. 4.8 cm

Answer: 2. 1.7 cm

Question 13. The angular magnification of a simple microscope can be increased by

1. Increasing the focal length of the lens
2. Increasing the size of the object
3. Increasing the aperture of the lens
4. Increasing the power of the lens

Answer: 4. Increasing the power of the lens

Question 14. The length of the tube of a microscope is 14 cm and its magnifying power for normal eye is 25. The focal length of the eyepiece is 5 cm. The distance of the object from the objective is

1. 2.4 cm
2. 2.5 cm
3. 3.6cm
4. 1.8 cm

Answer: 4. 1.8 cm

Question 15. Galileo’s telescope has an objective of a focal length of 100 cm and a magnifying power of 50. The distance between the two lenses in normal adjustment is

1. 106 cm
2. 102 cm
3. 92 cm
4. 98 cm

Answer: 4. 98 cm

Question 16. In Galileo’s telescope, the inverted image formed by its. objective serves as a virtual object for its eyepiece. If the eyepiece has to form an inverted and magnified image of the virtual object, the eyepiece has to be a concave lens and it must be so placed that the virtual object falls

1. Within F
2. Between F and 2F
3. At 2F
4. Beyond 2F

Answer: 1. Within F

Question 17. In an astronomical telescope, if the focal lengths of the objective and the eyepiece are f0 and fe respectively, then the imagination of this instrument is almost.
Answer:

1. f_o+f_e
2. f_o × f_e
3. \(\frac{f_o}{f_e}\)
4. ½ (f_o × f_e)

Answer: 3. \(\frac{f_o}{f_e}\)

Question 18. In the case of the normal focusing of an astronomical telescope, the final image is formed at

1. Focus of the eyepiece
2. Least distance of distinct vision
3. The focus of the objective
4. Infinity

Answer: 4. Focus on the objective

Question 19. The angular magnification of an astronomical telescope will be maximum if the focal lengths of the objective and the eyepiece are respectively

1. 1 m and 5 cm
2. 2 m and 6 cm
3. 3m and 4 cm
4. 4m and 3 cm

Answer: 4. 4m and 3 cm

Question 20. If the focal length of the eyepiece of a telescope is doubled, its magnifying power m will be

1. 2m
2. 3m
3. \(\frac{m}{2}\)
4. 4m

Answer: 3. \(\frac{m}{2}\)

Question 21. When we see an object, the image formed on the retina is

1. Real
2. Virtual
3. Erect
4. Inverted

Answer: 1 And 4

Question 22. In which of the following instruments is the final image erect?

1. Simple microscope
2. Compound microscope
3. Atronomical telescope
4. Galilean telescope

Answer: 1 And 4

Question 23. Mark the correct options.

1. If the far point increases, the power of the divergent less should be reduced
2. If the near point increases the power of the convergent lens should be reduced
3. If die far point is 1 in away from the eye, divergent less should be used
4. If the near print is 1 m away from the eye, a divergent lens should be used

Answer: 1 And 3

Question 24. The focal length of the objective of a compound microscope is f_o and its distance from the eyepiece is L An object is placed at a distance u from the objective. For proper working of the instrument which of the following options are suitable?

1. L<u
2. L>u-
3. f<L<2f
4. L>2f

Answer: 2 and 4

Question 25. A magnifying glass of focal length f used to see an object placed at a distance u from it forms the virtual image at the least distance of distinct vision D. Its magnifying power is given by

1. \(\frac{D}{f}\)
2. 1 – \(\frac{D}{f}\)
3. \(\frac{D}{u}\)
4. 1 + \(\frac{D}{f}\)

Answer: 3 And 4

Question 26. A planet is observed by an astronomical reflecting telescope having an objective of focal length 16 m and eyepiece of focal length 2 cm

1. The distance between the objective and the eyepiece is 16.02 m^-1
2. The angular magnification of the planet is 800
3. The image of the planet is inverted
4. The objective is larger than the eyepiece

Answer: 1,2,3 And 4

Question 27. A compound microscope has an objective lens of focal length 1 cm and an eyepiece of focal length 2.5 cm.

1. If a distinct image of an object situated at a distance of 1.05 cm from the objective is seen, the magnification will be

1. 320
2. 280
3. 220
4. 110

Answer: 3. 220

2. Under the above condition the distance between the two lenses will be

1. 21 cm
2. 32 cm
3. 2.27 cm
4. 23.27 cm

Answer: 4. 23.27 cm

Question 28. A figure divided into squares, each of size 1.0 mm² is being viewed at a distance of 9.0 cm through a magnifying lens of focal length 10 cm held close to the eye.

1. The magnification produced by the lens will be

1. 10
2. 5
3. 20
4. 25

Answer: 1. 10

2. The area of each square in the virtual image is

1. 1.5 cm²
2. 1 cm²
3. 1.8 cm²
4. 2 cm²

Answer: 2. 1 cm²

3. The angular magnification of the lens is equal to

1. 2.5
2. 3.8
3. 3.5
4. 4.2

Answer: 3. 3.5

Question 29. The focal lengths of the objective and the eyepiece of an astronomical telescope are 140 cm and 5 cm respectively.

1. The magnifying power of the telescope for viewing objects when the final image is formed at the least distance of distinct vision (25 cm) will be

1. 26.5
2. 30
3. 33.6
4. 40

Answer: 1. 26.5

2. When the telescope is in normal adjustment (i.e., the final image is formed at infinity), the magnifying power of the telescope will be

1. 24
2. 28
3. 32
4. 38

Answer: 2. 28

3. When the telescope is in normal adjustment, the separation between the objective and the eyepiece will be

1. 120 cm
2. 140 mm
3. 145 cm
4. 140 cm

Answer: 3. 145 cm

Question 30. In an astronomical telescope, the focal length of the objective is mad

1. Half that of the eyepiece
2. Equal to the eyepiece
3. Shorter than that of the eyepiece
4. Greater than that of the eyepiece

Answer: 4. Greater than that of the eyepiece

Question 31. The intermediate image formed by the objective of a compound microscope is

1. Real, Inverted mid magnified
2. Real erect and magnified
3. Virtual, erect, and magnified
4. Virtual, Inverted, and magnified

Answer: 1. Real, Inverted mid magnified

Question 32. An observer looks m ti distant tree of height 10 m with a of 20, To the observer the tree telescope of magnifying power appears

1. 10 times taller
2. 10 times nearer
3. 20 times taller
4. 20 times nearer

Answer: 3. 20 times taller

As the tree Is situated at a large distance, the rays coming from It may be taken as parallel. I fence, It can be considered that the tree Is situated at Infinity. In the case of focusing for Infinity, the final Image Is formed at Infinity. Now magnifying power

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

= 20

Hence, the tree appears 20 times taller

Question 33. If the focal length of the objective lens is Increased then the magnifying power of the

1. The microscope will Increase but that of the telescope decrease
2. Microscope and telescope both will decrease
3. Microscope and telescope both will Increase
4. The microscope will decrease but that of the telescope will Increase

Answer: 4. The microscope will decrease but that of the telescope will Increase

Question 34. An astronomical telescope has an objective and eyepiece of focal lengths of 40 cm and 4 cm respectively. To view 200 cm away from the objective, the lenses separated by a distance

1. 46 cm
2. 50 cm
3. 54cm
4. 37.3 cm

Answer: 3. 54cm

The focal length of the objective, f_o = 40 cm, the focal length of the eyepiece, f_e =4cm, and the distance of the objective

Objective, u_o = -200 an.

Let, the distance of the image from the objective = v_o

∴ \(\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o}\)

or, \(\frac{1}{v_o}+\frac{1}{200}=\frac{1}{40}\)

or, \(\frac{1}{v_0}=\frac{1}{40}-\frac{1}{200}\)

Or, v_o = 50 cm

If the distance between the eyepiece and the image formed by the tire objective is u_e, the distance of the tire objective from the image

Is v_e and the Image Is formed at Infinity

⇒ \(\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}\)

Or, \(\frac{1}{\infty}-\frac{1}{u_e}=\frac{1}{4}\)

u = – 4 cm

Therefore, the distance between the eyepiece and the objective, L =\(\left|v_o\right|+\left|u_{\mathrm{f}}\right|\)

= 50+ 4 = 54 cm

Question 35. A person has a near point at 60cm. The focal length of spectacles lenses to read at 22cm having glasses separated 2 cm from the eye is

1. 40 cm
2. 10cm
3. 20 cm
4. 30 cm

Answer: 4. 30 cm

Distance of the object from lens, u = -(22-2) = -20cm Image distance, v = -60cm

⇒ \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=-\frac{1}{60}+\frac{1}{20}=\frac{1}{30}\)

Hence, f = 30cm