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Class XII · Physics

Optics: NEET MCQ practice

413 practice questions on Optics, each with an explanation. Practise them chapter by chapter in the GyaanLoop app.

413

Practice questions

156

Easy

232

Medium

25

Hard

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What Optics covers

Topics in the NEET (UG) 2026–27 syllabus for this chapter.

  • Reflection and refraction of light at plane and spherical surfaces, mirror formula
  • Refraction through a prism, thin lens formula, magnification, power of a lens
  • Microscope and astronomical telescope- magnifying powers
  • Wavefront and Huygens principle, laws of reflection/refraction using Huygens principle
  • Interference, Young's double slit experiment, fringe width
  • Diffraction due to a single slit, width of central maximum
  • Polarisation, plane-polarised light, Brewster's law, uses of polaroids

Sample questions on Optics

Try each one, then open the answer and explanation.

Question 1 · Practice question

Different objects at different distances are seen by the eye. The parameter that remains constant is

  1. A.the focal length of the eye lens
  2. B.the object distance from the eye lens
  3. C.the radii of curvature of the eye lens
  4. D.the image distance from the eye lens
Show answer

Answer: D. the image distance from the eye lens

For clear vision, the eye forms images on the retina, which is at a fixed distance from the eye lens. While the focal length changes to focus objects at different distances, the image distance remains constant. Do not choose focal length—it adjusts via the ciliary muscles—or object distance, which obviously varies with the object's position.

Question 2 · Practice question

An under-water swimmer cannot see very clearly even in absolutely clear water because of

  1. A.absorption of light in water
  2. B.scattering of light in water
  3. C.reduction of speed of light in water
  4. D.change in the focal length of eye lens
Show answer

Answer: D. change in the focal length of eye lens

Underwater, the refractive index of water is close to that of the fluid inside the eye, so the cornea loses much of its bending power. This changes the effective focal length of the eye lens system, making the eye unable to focus sharply on objects. While speed of light reduces (option C) and scattering occurs (option B), they don't directly cause blurry vision in clear water — the main issue is the loss of refractive power at the cornea-water interface.

Question 3 · Practice question

The nearer point of hypermetropic eye is 40 cm. The lens to be used for its correction should have the power?

  1. A.+1.5 D
  2. B.-1.5 D
  3. C.+2.5 D
  4. D.+0.5 D
Show answer

Answer: A. +1.5 D

In hypermetropia, the near point is farther away than normal (25 cm). To correct it, a converging lens forms a virtual image of an object at 25 cm at the defective near point of 40 cm. Using the lens formula gives power P = 1/0.25 − 1/0.4 = +1.5 D. The tempting wrong option −1.5 D arises from mixing up myopia and hypermetropia signs.

Question 4 · Practice question

A microscope is focused 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?

  1. A.4.5 cm downward
  2. B.1 cm downward
  3. C.2 cm upward
  4. D.1 cm upward
Show answer

Answer: D. 1 cm upward

When a glass slab is placed over the mark, light from the mark bends upward, making it appear closer by an amount Δt = t(1 − 1/μ). Substituting t = 3 cm and μ = 1.5 gives an upward shift of 1 cm. To focus on this apparently raised mark, the microscope must be moved upward by the same 1 cm. A common mistake is to miscalculate the shift as 3 × (μ − 1) = 1.5 cm downward, leading to the wrong option (b).

Question 5 · Practice question

A compound microscope consists of an objective lens with focal length 1.0 cm and eye piece of focal length 2.0 cm and a tube length 20 cm the magnification will be

  1. A.100
  2. B.200
  3. C.250
  4. D.300
Show answer

Answer: C. 250

The magnifying power of a compound microscope in normal adjustment (final image at infinity) is given by M = (L/f₀) × (D/fₑ), where L is the tube length, f₀ and fₑ are focal lengths of objective and eyepiece, and D is the least distance of distinct vision (25 cm). Substituting the given values yields M = (20/1) × (25/2) = 250. The common mistake is using the formula M = (L/f₀) × (1 + D/fₑ) for the image at the near point, which would give 270, but since the question does not specify that, the conventional normal adjustment formula applies.

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