Ray Optics – Optical Instruments: NEET MCQs with Answers

15 free questions on Ray Optics – Optical Instruments for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Optical Instruments: Microscope and Astronomical Telescope (reflecting and refracting) and their magnifying powers.

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Question 1

Previous-year question
An astronomical refracting telescope will have large angular magnification and high angular resolution when it has an objective lens of [NEET 2018]
  1. small focal length and large diameter
  2. large focal length and small diameter
  3. small focal length and small diameter
  4. large focal length and large diameter
Show answer
Answer: (D) large focal length and large diameter
Explanation: Angular magnification f₀/fₑ calls for a long objective focal length, while resolving power D/1.22λ calls for a large objective diameter. Both requirements are met by a large-focal-length, large-diameter objective.

Question 2

Previous-year question
An astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance [NEET 2016]
  1. 37.3 cm
  2. 46.0 cm
  3. 50.0 cm
  4. 54.0 cm
Show answer
Answer: (D) 54.0 cm
Explanation: The objective images the object at v = 200 × 40/(200 − 40) = 50 cm. With the final image at infinity, the eyepiece must be one focal length beyond, so the separation is 50 + 4 = 54 cm.

Question 3

Previous-year question
An astronomical telescope of ten-fold angular magnification has a length of 44 cm. The focal length of the objective is [AIPMT 1997]
  1. 440 cm
  2. 44 cm
  3. 40 cm
  4. 4 cm
Show answer
Answer: (C) 40 cm
Explanation: For a telescope in normal adjustment, L = f₀ + f_e = 44 and M = f₀/f_e = 10, so f_e = f₀/10. Then f₀ + f₀/10 = 44, i.e. 11f₀/10 = 44, giving f₀ = 40 cm.

Question 4

Previous-year question
A microscope is focussed on a mark on a piece of paper and then a slab of glass of thickness 3 cm and refractive index 1.5 is placed over the mark. How should the microscope be moved to get the mark in focus again? [AIPMT 2006]
  1. 1 cm upward
  2. 4.5 cm downward
  3. 1 cm downward
  4. 2 cm upward
Show answer
Answer: (A) 1 cm upward
Explanation: The apparent depth of the mark is real depth/μ = 3/1.5 = 2 cm, so the mark appears raised by 3 − 2 = 1 cm. To refocus, the microscope must be moved up by the same amount, 1 cm.

Question 5

Previous-year question
The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm. The focal lengths of the lenses are [AIPMT 2012]
  1. 10 cm, 10 cm
  2. 15 cm, 5 cm
  3. 18 cm, 2 cm
  4. 11 cm, 9 cm
Show answer
Answer: (C) 18 cm, 2 cm
Explanation: For parallel rays, f₀ + f_e = 20 and f₀/f_e = 9, so f₀ = 9f_e. Substituting: 9f_e + f_e = 20, giving f_e = 2 cm and f₀ = 18 cm.

Question 6

Previous-year question
For a normal eye, the cornea of the eye provides a converging power of 40 D and the least converging power of the eye lens behind the cornea is 20 D. Using this information, the distance between the retina and the cornea–eye lens can be estimated to be [NEET 2013]
  1. 5 cm
  2. 2.5 cm
  3. 1.67 cm
  4. 1.5 cm
Show answer
Answer: (C) 1.67 cm
Explanation: The two elements are effectively in contact, so P_eq = P₁ + P₂ = 40 + 20 = 60 D. Hence f_eq = 100/60 cm = 1.67 cm. Since parallel light from a distant object must focus on the retina, this is the required distance.

Question 7

Previous-year question
If the focal length of the objective lens is increased, then the magnifying power of [AIPMT 2014]
  1. microscope will increase but that of telescope decrease
  2. microscope and telescope both will increase
  3. microscope and telescope both will decrease
  4. microscope will decrease but that of telescope will increase
Show answer
Answer: (D) microscope will decrease but that of telescope will increase
Explanation: For a compound microscope m = (L/f₀)(D/f_e), so m ∝ 1/f₀ and increasing f₀ reduces the magnification. For a telescope M = f₀/f_e, so M ∝ f₀ and increasing f₀ raises the magnification.
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Question 8

Previous-year question
In an astronomical telescope in normal adjustment, a straight black line of length L is drawn on the inside part of the objective lens. The eyepiece forms a real image of this line. The length of this image is I. The magnification of the telescope is [AIPMT 2015]
  1. L/I + 1
  2. L/I − 1
  3. (L + 1)/(L − 1)
  4. L/I
Show answer
Answer: (D) L/I
Explanation: In normal adjustment the objective and eyepiece are separated by f₀ + f_e, so for the eyepiece the line acts as an object at u = −(f₀ + f_e). Working through the lens equation gives f_e/f₀ = I/L, and since M = f₀/f_e, the magnification is M = L/I.

Question 9

Previous-year question
A person can see clearly objects only when they lie between 50 cm and 400 cm from his eyes. In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens the person has to use will be [NEET 2016]
  1. convex, +2.25 D
  2. concave, −0.25 D
  3. concave, −0.2 D
  4. convex, +0.15 D
Show answer
Answer: (B) concave, −0.25 D
Explanation: The lens must form an image of a distant object at the person's far point, so u = ∞ and v = −400 cm = −4 m. From 1/v − 1/u = 1/f, f = −4 m. Hence P = 1/f = −0.25 D, a concave (diverging) lens.

Question 10

Previous-year question
An astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance [NEET 2016]
  1. 46.0 cm
  2. 50.0 cm
  3. 54.0 cm
  4. 37.3 cm
Show answer
Answer: (C) 54.0 cm
Explanation: For the objective, u₀ = −200 cm and f₀ = 40 cm: 1/v = 1/40 − 1/200 = 4/200, so v = 50 cm. The image must lie at the focus of the eyepiece, so the separation is v + f_e = 50 + 4 = 54 cm.

Question 11

Previous-year question
A lens of large focal length and large aperture is best suited as an objective of an astronomical telescope, since [NEET 2021]
  1. a large aperture contributes to the quality and visibility of the images
  2. a large area of the objective ensures better light gathering power
  3. a large aperture provides a better resolution
  4. All of the above
Show answer
Answer: (D) All of the above
Explanation: The magnifying power M = f₀/f_e grows with the objective focal length, while a large aperture gathers more light (making faint objects visible), improves image quality and increases resolving power, since the limit of resolution 1.22λ/D falls as the diameter D rises. All three statements are therefore correct.

Question 12

Previous-year question
A small telescope has an objective of focal length 140 cm and an eyepiece of focal length 5.0 cm. The magnifying power of the telescope for viewing a distant object is [NEET 2024]
  1. 34
  2. 28
  3. 17
  4. 32
Show answer
Answer: (B) 28
Explanation: For a telescope in normal adjustment the magnifying power is M = f₀/f_e = 140/5 = 28.

Question 13

In a compound microscope, the objective has focal length f₀, the object is at a distance u from the objective, and the objective and the eyepiece are a distance L apart. For the instrument to work properly,
  1. L > 2f₀
  2. f₀ < L < 2f₀
  3. L = f₀
  4. L < u
Show answer
Answer: (A) L > 2f₀
Explanation: The object is placed just beyond the focus of the objective, so u is slightly greater than f₀, and a magnified real image forms at a distance v. A converging lens gives a magnified real image only when v exceeds 2f₀. This image must lie between the two lenses, so L > v > 2f₀.

Question 14

In which of these instruments, used in the usual way, is the final image erect with respect to the object?
  1. Newtonian reflecting telescope
  2. Compound microscope
  3. Simple microscope
  4. Refracting astronomical telescope
Show answer
Answer: (C) Simple microscope
Explanation: A simple microscope is a single converging lens with the object inside its focal length, which gives a virtual, erect, magnified image. In the other three instruments the objective or primary mirror forms an inverted intermediate image. The converging eyepiece leaves that image inverted.

Question 15

In a Cassegrain reflecting telescope, what happens to the light after it is reflected by the concave primary mirror?
  1. It comes to a focus inside the tube, where an observer in a small cage views it, with no incoming light blocked
  2. A converging lens at the centre of the primary mirror brings it to a focus, which adds to the light collected by the mirror
  3. A plane mirror turns it through 90° out of the side of the tube, which removes all obstruction of the incoming light
  4. A convex secondary mirror sends it through a hole in the primary mirror to the eyepiece, giving a large focal length in a short telescope
Show answer
Answer: (D) A convex secondary mirror sends it through a hole in the primary mirror to the eyepiece, giving a large focal length in a short telescope
Explanation: A mirror objective focuses light inside the tube, so an eyepiece placed there blocks some light. The Cassegrain arrangement deflects the converging light with a convex secondary mirror, and it passes through a hole in the primary. Folding the light path in this way gives a large focal length in a short telescope.

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