Ray Optics and Optical Instruments NEET PYQReal NEET & AIPMT 2006–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Ray Optics and Optical Instruments question from the NEET & AIPMT papers on this site (2006–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Ray Optics and Optical Instruments previous year questions
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Q1NEET 2021
A convex lens 'A' of focal length 20 cm and a concave lens 'B' of focal length 5 cm are kept along the same axis with a distance 'd' between them. If a parallel beam of light falling on 'A' leaves 'B' as a parallel beam, then the distance 'd' in cm will be :
Not quite — the answer is B.
For the beam to emerge parallel, the second focal point of the convex lens must coincide with the focal point of the concave lens on the far side. Since the concave lens's focus lies on the incoming side at 5 cm, d = 20 − 5 = 15 cm. Adding the focal lengths (25 cm) would be right only for two convex lenses.
Q2NEET 2020 · Q140
A ray is incident at an angle of incidence i on one surface of a small angle prism (with angle of prism A) and emerges normally from the opposite surface. If the refractive index of the material of the prism is μ, then the angle of incidence is nearly equal to :
Not quite — the answer is C.
Emerging normally means r₂ = 0, so r₁ = A − r₂ = A. For small angles Snell's law sin i = μ sin r₁ becomes i ≈ μA. Dividing by μ instead of multiplying confuses the direction of refraction.
Q3NEET 2019 · Q11
Two similar thin equi-convex lenses, of focal length f each, are kept coaxially in contact with each other such that the focal length of the combination is F₁. When the space between the two lenses is filled with glycerin (which has the same refractive index (μ = 1.5) as that of glass) then the equivalent focal length is F₂. The ratio F₁ : F₂ will be:
Not quite — the answer is B.
For an equiconvex glass lens (μ = 1.5) f = R, so F₁ = f/2. The glycerin film is a concave lens with 1/f = 0.5(−2/R) = −1/f. Then 1/F₂ = 1/f + 1/f − 1/f = 1/f, so F₂ = f and F₁ : F₂ = 1 : 2. The ratio 2 : 1 inverts this.
Q4NEET 2019 · Q28
In total internal reflection when the angle of incidence is equal to the critical angle for the pair of media in contact, what will be angle of refraction?
Not quite — the answer is D.
The critical angle is defined as the angle of incidence for which the refracted ray grazes the interface, i.e. the angle of refraction is 90°. Beyond it, total internal reflection occurs.
Q5NEET 2018 · Q10
An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of
Not quite — the answer is A.
Magnification m = f₀/fₑ grows with the objective's focal length, and resolving power ∝ D/λ grows with its aperture. So the objective should have both a large focal length and a large diameter; a small aperture would worsen resolution.
Q6NEET 2018 · Q19
The refractive index of the material of a prism is √2 and the angle of the prism is 30°. One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is
Not quite — the answer is B.
To retrace its path the ray must hit the silvered face normally, so r₂ = 0 and r₁ = A = 30°. By Snell's law sin i = √2 sin 30° = 1/√2, so i = 45°. Zero incidence would make the ray strike the silvered face at 30°, not normally.
Q7NEET 2018 · Q20
An object is placed at a distance of 40 cm from a concave mirror of focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be
Not quite — the answer is B.
With u = −40 cm, f = −15 cm: 1/v = −1/15 + 1/40 gives v = −24 cm. With u = −20 cm: 1/v = −1/15 + 1/20 gives v = −60 cm. The image moves from 24 cm to 60 cm, i.e. 36 cm away from the mirror.
Q8NEET 2017 · Q138
The ratio of resolving powers of an optical microscope for two wavelengths λ₁ = 4000 Å and λ₂ = 6000 Å is:
Not quite — the answer is C.
Resolving power of a microscope is 2n sin θ/λ, i.e. inversely proportional to wavelength. RP₁ : RP₂ = λ₂ : λ₁ = 6000 : 4000 = 3 : 2. Squaring or cubing the ratio (9 : 4, 8 : 27) has no basis.
Q9NEET 2017 · Q144
A beam of light from a source L is incident normally on a plane mirror fixed at a certain distance x from the source. The beam is reflected back as a spot on a scale placed just above the source L. When the mirror is rotated through a small angle θ, the spot of the light is found to move through a distance y on the scale. The angle θ is given by:
Not quite — the answer is A.
Rotating a mirror by θ turns the reflected ray by 2θ. For a small angle the spot moves y = x·2θ on the scale at distance x, so θ = y/2x. Writing θ = y/x forgets the doubling of the reflected beam's rotation.
Q10NEET 2017 · Q160
A thin prism having refracting angle 10° is made of glass of refractive index 1.42. This prism is combined with another thin prism of glass of refractive index 1.7. This combination produces dispersion without deviation. The refracting angle of second prism should be:
Not quite — the answer is B.
For zero net deviation (μ₁ − 1)A₁ = (μ₂ − 1)A₂, so 0.42 × 10° = 0.7 × A₂ and A₂ = 6°. A smaller-angle prism of higher index cancels the deviation of the first.
Q11NEET 2017 (Gujarati paper) · Q137
If the angle of a prism is 60° and angle of minimum deviation is 40°, then the angle of refraction will be:
Not quite — the answer is A.
At minimum deviation the ray passes symmetrically, so r₁ = r₂ and r₁ + r₂ = A. Hence r = A/2 = 60°/2 = 30°, whatever the deviation. 40° belongs to the angle of incidence calculation, i = (A + δ)/2 = 50°, not to r.
Q12NEET 2017 (Gujarati paper) · Q149
A light beam is incident on a denser medium whose refractive index is 1.414 at an angle of incidence 45°. Find the ratio of width of refracted beam in a medium to the width of the incident beam in air.
Not quite — the answer is D.
sin r = sin 45°/1.414 = 0.5, so r = 30°. A beam crossing a boundary strip of length x has width x cos i in air and x cos r in the medium, so the ratio is cos 30°/cos 45° = (√3/2)/(1/√2) = √3 : √2. The inverse ratio √2 : √3 comes from swapping the two angles.
Q13NEET 2016 Phase 2 · Q34
Two identical glass (μ_g = 3/2) equiconvex lenses of focal length f each are kept in contact. The space between the two lenses is filled with water (μ_w = 4/3). The focal length of the combination is
Not quite — the answer is D.
For each glass lens 1/f = (0.5)(2/R), so R = f. The water forms a concave lens with 1/f_w = (1/3)(−2/R) = −2/(3f). Combined: 1/F = 2/f − 2/(3f) = 4/(3f), so F = 3f/4. Ignoring the water lens gives f/2, and treating it as converging gives a shorter focal length.
Q14NEET 2016 Phase 2 · Q35
An air bubble in a glass slab with refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness (in cm) of the slab is
Not quite — the answer is C.
Real depth = μ × apparent depth, so the bubble is 1.5 × 5 = 7.5 cm from one face and 1.5 × 3 = 4.5 cm from the other. Thickness = 7.5 + 4.5 = 12 cm. Just adding the apparent depths gives 8 cm.
Q15NEET 2016 Phase 2 · Q37
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
Not quite — the answer is B.
To see distant objects the lens must form the image of an object at infinity at the far point, 400 cm, so f = −400 cm = −4 m. Power P = 1/f = −0.25 D, a concave lens. A convex lens would correct the near point, not the far point.
Q16NEET 2016 Phase 1 · Q22
The angle of incidence for a ray of light at a refracting surface of a prism is 45°. The angle of prism is 60°. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are:
Not quite — the answer is B.
At minimum deviation i = e, so δ = 2i − A = 90° − 60° = 30°, and r = A/2 = 30°. μ = sin45°/sin30° = √2. A refractive index of 1/√2 (less than 1) is impossible for a prism in air.
Q17NEET 2016 Phase 1 · Q32
A 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:
Not quite — the answer is D.
Objective: 1/v − 1/(−200) = 1/40 gives v = 50 cm. For normal adjustment the image must be at the focus of the eyepiece, so separation = 50 + 4 = 54 cm. 44 cm (fo + fe) applies only to an object at infinity, and 50 cm ignores the eyepiece.
Q18NEET 2016 Phase 1 · Q44
Match the corresponding entries of column 1 with column 2. [Where m is the magnification produced by the mirror] Column 1: (A) m = −2, (B) m = −1/2, (C) m = +2, (D) m = +1/2 Column 2: (a) Convex mirror, (b) Concave mirror, (c) Real image, (d) Virtual image
Not quite — the answer is A.
A negative magnification means an inverted real image, which only a concave mirror forms, so m = −2 and m = −1/2 are both concave mirror with real image. m = +2 is an enlarged erect virtual image, possible only with a concave mirror, while m = +1/2 is a diminished erect virtual image, typical of a convex mirror. The matching that pairs a convex mirror with a real image is impossible.
Q19AIPMT 2015
The refracting angle of a prism is A, and refractive index of the material of the prism is cot (A/2). The angle of minimum deviation is:
Not quite — the answer is C.
μ = sin((A + δm)/2)/sin(A/2) = cot(A/2) = cos(A/2)/sin(A/2). So sin((A + δm)/2) = cos(A/2) = sin(90° − A/2), giving (A + δm)/2 = 90° − A/2 and δm = 180° − 2A.
Q20AIPMT 2015
Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index 1.7. The focal length of the combination is:
Not quite — the answer is D.
Each plano-convex glass lens: 1/f = (1.5 − 1)/20 = 1/40 cm⁻¹. The oil forms a biconcave lens: 1/f = (1.7 − 1)(−1/20 − 1/20) = −0.07 cm⁻¹. Total 1/F = 2/40 − 0.07 = −0.02 cm⁻¹, so F = −50 cm.
Q21AIPMT 2014
If the focal length of objective lens is increased then magnifying power of:
Not quite — the answer is C.
For a compound microscope M ≈ (L/f₀)(D/fₑ), so a larger objective focal length lowers M. For a telescope M = f₀/fₑ, so it rises; the two instruments respond oppositely.
Q22AIPMT 2014
The angle of a prism is 'A'. One of its refracting surfaces is silvered. Light rays falling at an angle of incidence 2A on the first surface returns back through the same path after suffering reflection at the silvered surface. The refractive index μ, of the prism is:
Not quite — the answer is A.
To return along its path the ray must strike the silvered face normally, so the refraction angle at the first face equals the prism angle, r = A. Then μ = sin 2A/sin A = 2 cos A.
Q23NEET 2013
A plano convex lens fits exactly into a plano concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices μ₁ and μ₂ and R is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is
Not quite — the answer is C.
Plano-convex: 1/f₁ = (μ₁ − 1)/R; plano-concave of the same curvature: 1/f₂ = −(μ₂ − 1)/R. In contact: 1/f = (μ₁ − μ₂)/R, so f = R/(μ₁ − μ₂). A factor of 2 would appear only for double-curved lenses.
Q24NEET 2013
For a normal eye, the cornea of 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
Not quite — the answer is C.
For a relaxed normal eye viewing a distant object, the total power 40 + 20 = 60 D must focus the image on the retina. f = 1/60 m = 100/60 cm ≈ 1.67 cm, which is the cornea–retina distance. Using only the cornea power (40 D) gives 2.5 cm.
Q25AIPMT 2012 Mains
For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index :
Not quite — the answer is B.
μ = sin((A + δm)/2)/sin(A/2); with δm = A this becomes μ = sin A/sin(A/2) = 2cos(A/2). For a real emergent ray A/2 ranges between 0 and 45°, so μ lies between √2 and 2.
Q26AIPMT 2012 Mains
A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is :
Not quite — the answer is D.
The near end at 20 cm = 2f images at 20 cm. For the far end, u = −30 cm: 1/v = 1/f − 1/u = −1/10 + 1/30, so v = −15 cm. The image extends from 15 cm to 20 cm, a length of 5 cm.
Q27AIPMT 2012 Prelims
When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid, it acts as a plane sheet of glass. This implies that the liquid must have refractive index
Not quite — the answer is B.
By the lens maker's formula 1/f = (μ_g/μ_l − 1)(1/R₁ − 1/R₂). The lens behaves like a plane sheet (no power) only when μ_g/μ_l = 1, i.e. the liquid has the same refractive index as glass, 1.47. A liquid of smaller index would leave it a converging lens.
Q28AIPMT 2012 Prelims
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 length of lenses are
Not quite — the answer is D.
For normal adjustment, magnifying power = f₀/fₑ = 9 and tube length = f₀ + fₑ = 20 cm. Solving, fₑ = 2 cm and f₀ = 18 cm. 15 cm and 5 cm add to 20 cm but give a magnification of only 3.
Q29AIPMT 2012 Prelims
A ray of light is incident at an angle of incidence i on one face of a prism of angle A (assumed to be small) and emerges normally from the opposite face. If the refractive index of the prism is μ, the angle of incidence i is nearly equal to
Not quite — the answer is B.
Normal emergence means r₂ = 0, so r₁ = A − r₂ = A. Snell's law at the first face: sin i = μ sin A. For small angles, i ≈ μA. A/μ would come from inverting Snell's law.
Q30AIPMT 2011 Mains
A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point 15 cm from the lens on the opposite side. If the lens is removed the point where the rays meet will move 5 cm closer to the lens. The focal length of the lens is
Not quite — the answer is A.
Without the lens the rays meet 10 cm behind the lens position, so that point is a virtual object at u = +10 cm; with the lens the image is at v = +15 cm. 1/f = 1/v − 1/u = 1/15 − 1/10 = −1/30, so f = −30 cm. The negative sign confirms a diverging lens, ruling out the positive values.
Q31AIPMT 2011 Mains
A thin prism of angle 15° made of glass of refractive index μ₁ = 1.5 is combined with another prism of glass of refractive index μ₂ = 1.75. The combination of the prisms produces dispersion without deviation. The angle of the second prism should be
Not quite — the answer is D.
For no net deviation with thin prisms, (μ₁ − 1)A₁ = (μ₂ − 1)A₂, so 0.5 × 15° = 0.75 × A₂ and A₂ = 10°. Using the ratio of refractive indices directly instead of (μ − 1) leads to wrong angles.
Q32AIPMT 2011 Prelims
Which of the following is not due to total internal reflection?
Not quite — the answer is C.
The pond looking shallower than it is comes from ordinary refraction at the water surface, not total internal reflection. Sparkle of diamond, light guidance in optical fibres and mirages all rely on total internal reflection.
Q33AIPMT 2010 Mains · Q68
The speed of light in media M₁ and M₂ is 1.5 × 10⁸ m/s and 2.0 × 10⁸ m/s respectively. A ray of light enters from medium M₁ to M₂ at an incidence angle i. If the ray suffers total internal reflection, the value of i is
Not quite — the answer is C.
Refractive indices are μ₁ = c/v₁ = 2 and μ₂ = c/v₂ = 1.5. The critical angle satisfies sin C = μ₂/μ₁ = 3/4, so total internal reflection needs i equal to or greater than sin⁻¹(3/4). The value sin⁻¹(2/3) is obtained by wrongly using 1/μ₁ ratios; TIR never occurs for angles smaller than the critical angle.
Q34AIPMT 2010 Mains · Q69
A ray of light is incident on a 60º prism at the minimum deviation position. The angle of refraction at the first face (i.e., incident face) of the prism is
Not quite — the answer is B.
At minimum deviation the ray passes symmetrically, so r₁ = r₂ and r₁ + r₂ = A. Hence r₁ = A/2 = 60°/2 = 30°. Taking 60° confuses the refraction angle with the prism angle itself.
Q35AIPMT 2010 Prelims · Q23
A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter d/2 in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively
Not quite — the answer is D.
Covering part of a lens does not change its focal length, which depends only on curvature and refractive index. The light gathered is ∝ area: the covered central part has diameter d/2, i.e. 1/4 of the area, so intensity becomes I − I/4 = 3I/4. I/4 would be the share of the covered part, not what remains.
Q36AIPMT 2010 Prelims · Q24
A ray of light travelling in a transparent medium of refractive index μ, falls on a surface separating the medium from air at an angle of incidence of 45º. For which of the following value of μ the ray can undergo total internal reflection?
Not quite — the answer is D.
TIR needs the incidence angle to exceed the critical angle: sin 45° > 1/μ, so μ > √2 ≈ 1.414. Only μ = 1.50 exceeds this; 1.40 is just below.
Q37AIPMT 2008 Prelims · Q23
Two thin lenses of focal lengths f₁ and f₂ are in contact and coaxial. The power of the combination is
Not quite — the answer is A.
For thin lenses in contact, powers add: P = 1/f₁ + 1/f₂ = (f₁ + f₂)/(f₁f₂). (f₁ + f₂)/2 has dimensions of length, so it cannot be a power.
Q38AIPMT 2008 Prelims · Q24
A boy is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length 10 cm. The diameter of the Sun is 1.39 × 10⁹ m and its mean distance from the earth is 1.5 × 10¹¹ m. What is the diameter of the Sun's image on the paper?
Not quite — the answer is B.
The image forms in the focal plane, and the Sun and its image subtend the same angle at the lens: d = f × D/u = 0.10 × 1.39 × 10⁹/1.5 × 10¹¹ ≈ 9.3 × 10⁻⁴ m, closest to 9.2 × 10⁻⁴ m. 6.5 × 10⁻⁴ m would come from wrongly halving or mis-scaling the angular size.
Q39AIPMT 2007 Prelims · Q13
The frequency of a light wave in a material is 2 × 10¹⁴ Hz and wavelength is 5000 Å. The refractive index of material will be
Not quite — the answer is B.
Speed in the material v = fλ = 2 × 10¹⁴ × 5 × 10⁻⁷ = 1 × 10⁸ m/s, so n = c/v = 3 × 10⁸/1 × 10⁸ = 3.00. 1.50 comes from a slip in the power of ten.
Q40AIPMT 2006 Prelims · Q18
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 ?
Not quite — the answer is A.
A slab of thickness t and refractive index μ makes the mark appear raised by t(1 − 1/μ) = 3(1 − 1/1.5) = 1 cm. Since the image of the mark is now 1 cm higher, the microscope must be raised by 1 cm. Moving it down would take it further out of focus.
Q41AIPMT 2006 Prelims · Q42
A convex lens and a concave lens, each having same focal length of 25 cm, are put in contact to form a combination of lenses. The power in diopters of the combination is :
Not quite — the answer is D.
The powers are +1/0.25 = +4 D for the convex lens and −4 D for the concave lens. In contact the total power is 4 − 4 = 0, so the combination acts like a plane glass plate. Zero power means infinite focal length, not infinite power.
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