This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Optical Instruments”.

1. The inability of a lens to form a white image of a white object is known as ________

a) Spherical Aberration

b) Chromatic Aberration

c) Monochromatic Aberration

d) Coma

View Answer

Explanation: If a lens forms colored images of an object with white light, it is known as chromatic aberration. Because since that the refractive index of the material of the lens is different for different wavelengths of light.

2. What is the ratio of the focal lengths of the two Plano-convex lenses in Huygens’s Eyepiece?

a) 2:1

b) 3:1

c) 3:2

d) 4:3

View Answer

Explanation: The focal lengths of the two Plano-convex lenses in Huygens’s Eyepiece are in the ration 3:1(3f and f) and the distance between them is equal to 2f. The focal length and positions of the two lenses are such that the eyepiece is achromatic and free from spherical aberration.

3. In which of the following instruments, the objective has a large focal length and a very large eyepiece?

a) A simple microscope

b) A Compound microscope

c) Telescope

d) Interferometer

View Answer

Explanation: In telescopes, it is desired to provide angular magnification of distant objects. A very large focal length and the even larger eyepiece is used so that the light from a distant object enters the objective and the image formed is clear.

4. X is an optical defect due to which a comet-like image is formed instead of a point image. X is ___________

a) Coma

b) Astigmatism

c) Curvature

d) Distortion

View Answer

Explanation: Coma is a type of monochromatic aberration. In this, a comet-like image is formed instead of a point image, of a point object situated away from the lens. It can be reduced by using the aplanatic lens.

5. Huygens’s eyepiece is also known as __________

a) Spherical Eyepiece

b) Positive Eyepiece

c) Negative Eyepiece

d) Double Eyepiece

View Answer

Explanation: In Huygens’s eyepiece, the real inverted image formed by the objective of the microscope lies behind the field lens and that image is virtual. Due to this, Huygens’s eyepiece is also known as Negative Eyepiece. This eyepiece cannot be used to examine a real image formed by the objective.

6. Which of the following is not a part of an Electron Microscope?

a) Electron Gun

b) Objective

c) Magnetic lens

d) Fluorescent Screen

View Answer

Explanation: The electron microscope works on the principle that a beam of electrons exhibit wave nature and they can be focused by suitable electric and magnetic fields. Thus, electron microscope has an electron gun for producing electrons, a Magnetic lens to focus the electron beam and the fluorescent screen to obtain the image.

7. The condition in which lines in one direction appear to be well focused while those in perpendicular direction appear distorted is known as ___________

a) Presbyopia

b) Myopia

c) Hypermetropia

d) Astigmatism

View Answer

Explanation: Astigmatism occurs when the cornea is not spherical. Due to this, lines in one direction seems to be focused while in the perpendicular direction looks distorted. It can be corrected by the use of cylindrical lenses.

8. A thin converging lens and a thin diverging lens of each focal length 10 cm are placed coaxially 5cm apart. What will be the focal length of the combination?

a) +10cm

b) -10cm

c) +20cm

d) -20cm

View Answer

Explanation: We know, f

_{1}=10 cm, f

_{2}= -10 cm, d = 5 cm

\(\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1f_2}\)

= 1/10 – 1/10 + 5/100

f = + 20 cm.

9. Two lenses of focal length 8 cm and 6 cm are placed a certain distance apart. If they form an achromatic combination, the separation between them is ___________

a) 5cm

b) 6cm

c) 7cm

d) 8cm

View Answer

Explanation: For an achromatic combination, the separation x is given by,

x = (f1 + f2)/2

As we know, f1 = 8cm and f2 = 6cm

x = 14/2

x = 7 cm.

10. The effective focal length of Ramsden’s eyepiece is 3cm. The focal length of a single lens is ___________

a) 3cm

b) 4cm

c) 5cm

d) 6cm

View Answer

Explanation: Here, F = 3 cm, f =? d =?

In Ramsden’s eyepiece, F = f and d = 2f/3

Therefore, applying the formula

\(\frac{1}{F}=\frac{1}{F}+\frac{1}{f}-\frac{d}{Ff}\)

F = 3f/4

f = 4F/3

f = 4 X 3 /3

f = 4 cm.

11. Find the position of the image in the following lens combination:

a) 30 cm to the right of the third lens

b) 20 cm to the right of the third lens

c) 30 cm to the left of the third lens

d) 20 cm to the left of the third lens

View Answer

Explanation: Image by the first lens,

\(\frac{1}{V_1}-\frac{1}{U_1}=\frac{1}{f_1} \)

\(\frac{1}{V_1}-\frac{1}{-30}=\frac{1}{10}\)

v

_{1}= 15 cm

For the second lens,

\(\frac{1}{V_2}\frac{-1}{-10}=\frac{1}{10}\)

V

_{2}= ∞

For the third,

\(\frac{1}{V_3}\frac{-1}{\infty}=\frac{1}{30}\)

V

_{3}= 30 cm.

12. The far point of a myopic person is 40 cm. What should be the power of the lens that he must use to see clearly?

a) -0.4 D

b) +0.4 D

c) -2.5 D

d) +2.5 D

View Answer

Explanation: Myopia is the defect when a person can’t see distant object clearly. To see clearly, the person should use a concave lens of focal length -40 cm.

Hence, the power of the lens = 100/-40

= -2.5 Diopters.

13. If the magnification of a lens is dependent on the distance from the principal axis, the aberration that arises is called __________

a) Coma

b) Astigmatism

c) Curvature

d) Distortion

View Answer

Explanation: Distortion is the defect that arises when the images of equal parts of an object will not be of the same length. In such a case, the magnification of the lens is dependent on the distance from the principal axis.

14. The refractive index of a Plano-convex lens is 1.6 for violet and 1.5 for red light. The radius of curvature is 0.20 cm. The separation between violet and red foci of the lens is ____________

a) 0.05 cm

b) 0.06 cm

c) 0.07 cm

d) 0.08 cm

View Answer

Explanation: Here, μ

_{v}= 1.6, μ

_{r}= 1.5, R

_{1}= 0.20 m, R

_{2}= ∞

Using lens maker’s formula: \(\frac{1}{F_v}=(μ_v-1)(\frac{1}{R_1}+\frac{1}{R_2})\)

F

_{v}= 0.33 m

Similarly solving for red color: \(\frac{1}{F_r}=(μ_r-1)(\frac{1}{R_1}+\frac{1}{R_2})\)

F

_{r}= 0.40 m

Separation between red and violet, F

_{r}–F

_{v}= 0.40 – 0.33

= 0.07 m.

15. The component in an optical instrument used to increase the angular object field and to minimize aberrations is called as ___________

a) Objective lens

b) Eye lens

c) Field Lens

d) Plano-concave lens

View Answer

Explanation: Field lens is placed between the objective and the eye lens. It increases the field of view and brightness of the image. It helps in reducing the aberrations in the lens. It brings the center of the exit pupil near the eye lens. Together, the field lens and the eye lens constitute an eyepiece or ocular.

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