Class 12 Physics MCQ – Optical Instruments

This set of Class 12 Physics Chapter 9 Multiple Choice Questions & Answers (MCQs) focuses on “Optical Instruments”.

1. A boy uses spectacles of focal length -50 cm. Name the defect of the vision he is suffering from.
a) Astigmatism
b) Hypermetropia
c) Myopia
d) Presbyopia
View Answer

Answer: c
Explanation: As the focal length is negative, the lens used is concave. When a person is prescribed a concave lens, then the person is considered to be suffering from myopic. Therefore, the boy is suffering from myopia.

2. How much intensity of the image is increased if the diameter of the objective of a telescope is doubled?
a) Two times
b) Four times
c) Eight times
d) Sixteen times
View Answer

Answer: b
Explanation: On doubling the diameter, the area of the objective increases four times. Its light-gathering power increases four times. The brightness of the image also increases four times. Therefore, the intensity of the image increases by four.

3. A fly is sitting on the objective of a telescope. What will be its effect on the final image of the distant object?
a) Reduces
b) Increases
c) Remains constant
d) Indefinite
View Answer

Answer: a
Explanation: The fly will not be seen in the final image. But the intensity of the final image gets reduced because light will not enter the telescope from that part of the object where the fly is sitting. Therefore, a reduction effect will be observed in the final image of the fly.

4. Two lenses of focal lengths 5 cm and 50 cm are to be used for making a telescope. Which lens will you use for the objective?
a) Both
b) Neither
c) 5 cm
d) 50 cm
View Answer

Answer: d
Explanation: In a telescope, the objective should have a large focal length than the eyepiece. So the lens of 50 cm focal length will be used as the objective. So, the smaller one of the two, i.e. the 5 cm focal length lens will be used as the eyepiece.

5. Which of the following is used to increase the range of a telescope?
a) Increasing the focal length
b) Decreasing the focal length
c) Increasing the diameter of the objective
d) Decreasing the diameter of the objective
View Answer

Answer: c
Explanation: The light-gathering power of the objective will increase and even faint objects will become visible. This is done by increasing the diameter of the objective. So, by increasing the diameter of the objective, the range of the telescope can be increased.
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6. The upper part of the bi-focal lens is a convex lens while its lower part is a concave lens.
a) True
b) False
View Answer

Answer: b
Explanation: No, this is a false statement. The upper part of the bi-focal lens should be a concave lens, whereas the one used for distant vision as its lower part should be a convex lens. Such a bi-focal lens is used for reading purposes.

7. ‘X’ can see objects in the ultra-violet light while human beings cannot do so. Identify X.
a) Penguin
b) Bees
c) Ant
d) Tiger
View Answer

Answer: b
Explanation: Bees have some retinal cones that are sensitive to ultra-violet light, so they can see objects in ultra-violet light. Humans are ultra-violet blind. Therefore, a human being cannot see UV light, whereas bees can.

8. How does the magnifying power of a telescope change on increasing the diameter of its objective?
a) Independent
b) Doubled
c) Halved
d) Becomes zero
View Answer

Answer: a
Explanation: Magnifying power of a telescope is given as:
m = \(\frac {f_0}{f_e}\)
It is independent of the aperture of the object. So the magnifying power is not affected when the diameter of the objective is increased.

9. What focal length should the reading spectacles have for a person for whom the least distance of distinct vision is 50 cm?
a) + 25 cm
b) – 25 cm
c) + 50 cm
d) – 50 cm
View Answer

Answer: c
Explanation: By thin lens formula,
\(\frac {1}{f} = \frac {1}{v} – \frac {1}{u}\)
\(\frac {1}{f} = \frac {1}{(-50)} – \frac {1}{(-25)}\)
f = +50 cm.

10. An object is to be seen through a simple microscope of power 10 D. Where should the object be placed to reduce maximum angular magnification? The least distance for distinct vision is 25 cm.
a) + 7.1 cm
b) – 7.1 cm
c) + 25 cm
d) – 25 cm
View Answer

Answer: b
Explanation: Angular magnification is maximum when the final image is formed at the near point.
\(\frac {1}{u} = \frac {1}{v} – \frac {1}{f}\)
\(\frac {1}{u} = ( \frac {-1}{25} ) – ( \frac {1}{10} ) \)
u = -7.1 cm.
Therefore, the object should be placed 7.1 cm before the lens to avoid angular magnification.

Sanfoundry Global Education & Learning Series – Physics – Class 12.

To practice all chapters and topics of class 12 Physics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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