25 questions in 25 minutes on the NTA computer-based test interface, all from Ray Optics and Optical Instruments. +4 / -1 marking, instant score, full solutions. Free, no login.
A chapter test is the fastest honest check on whether a chapter has actually landed. This one draws 25 questions from Ray Optics and Optical Instruments, of which 18 are real NEET previous-year questions and 6 sit at the harder end of the bank. Sit it in one 25-minute block, the way you would in the hall.
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Duration: 180 minutes · 180 questions · 720 marks maximum
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NEET CBT mock test: common questions
How many questions are in this Ray Optics and Optical Instruments mock test?
25 questions in 25 minutes, marked +4 for a correct answer and -1 for a wrong one, exactly like the real paper. 18 of them are NEET previous-year questions.
Is this the same interface as the real NEET CBT?
Yes. On-screen countdown, colour-coded question palette, Save & Next, Clear Response and Mark for Review, then an instant score with full solutions.
Should I take this before or after revising Ray Optics and Optical Instruments?
Both, and that is the point. Take it cold to find out what you actually do not know, revise from the chapter notes, then retake it to confirm the gap closed.
All 25 questions with answers and solutions
The complete paper in text form, with the correct option and a worked explanation for every question. Sit the timed test above first: the score is only worth having if you earn it. Then open this to revise, or to re-read a question you got wrong without replaying the whole paper.
Show all 25 questions with answers and solutions
Physics
- Q1A lens of large focal length and large aperture is best suited as an objective of an astronomical telescope since:
- a large aperture provides a better resolution.
- All the above.
- a large area of the objective ensures better light gathering power.
- a large aperture contributes to the quality and visibility of the images.
Answer: (B) All the above.
D. Tests: why telescope objectives are built with large aperture and large focal length. Why D: A large aperture gathers more light (area scales as the square of the diameter), so faint stars appear brighter; it also reduces the diffraction limit θ≈(1.22λ)/(D), giving better resolution; together these improve image quality and visibility. Since A, B and C are all true benefits, the best answer is all of the above. The large focal length separately boosts magnification m=(fₒ)/(fₑ). Why not A: true but incomplete, image quality is only one consequence of a big aperture [misconception A: “the first true statement must be the answer”] Why not B: true but incomplete, light gathering is only part of the story [misconception B: “aperture only affects brightness”] Why not C: true but incomplete, resolution is only part of the story [misconception C: “aperture only affects resolution”] Remember: big objective aperture = brighter AND sharper; big objective focal length = more magnification.
- Q2An air bubble in a glass slab (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:
- 12
- 8
- 16
- 10
Answer: (A) 12
For the two faces: (x)/(μ)+(t-x)/(μ)=3+5, i.e. (t)/(μ)=8. So t=8μ=8×1.5=12 cm.
- Q3An object is mounted on a wall. Its image of equal size is to be obtained on a parallel wall with the help of a convex lens placed between these walls. The lens is kept at distance x in front of the second wall. The required focal length of the lens will be:
- (x)/(2)
- less than (x)/(4)
- (x)/(4)
- more than (x)/(4) but less than (x)/(2)
Answer: (A) (x)/(2)
C. Tests: the equal-size real image condition for a convex lens. Why C: An equal-size real image means |m|=1, which happens only when the object is at 2f and the image is at 2f on the other side. The image forms on the second wall, so the lens-to-image distance is x, giving 2f=x, i.e. f=(x)/(2). Why not A: less than (x)/(4) misapplies the displacement-method condition f≤(D)/(4), and even that uses the wall-to-wall separation D, not the lens-to-wall distance x [misconception A: “f = separation/4 governs every two-wall setup”] Why not B: no lens rule places f strictly between (x)/(4) and (x)/(2); it hedges between two half-remembered formulas [misconception B: “averaging remembered results”] Why not D: (x)/(4) again confuses x with the full object-to-image distance 4f [misconception D: “x is the wall-to-wall distance”] Remember: unit magnification puts object and image at 2f each; here the image leg alone is x=2f.
- Q4A convex lens forms a virtual, erect and magnified image when the object is placed:
- between F and 2F
- at 2F
- between the lens and F
- beyond 2F
Answer: (C) between the lens and F
When the object is inside the focal length (between lens and F), a convex lens forms a virtual, erect, magnified image – the magnifying glass action.
- Q5Assertion: The objective of an astronomical telescope has a large aperture. Reason: A large aperture gathers more light and improves resolving power.
- Assertion false, reason true
- Both assertion and reason are true and reason correctly explains assertion
- Assertion true, reason false
- Both are true but reason does not explain assertion
Answer: (B) Both assertion and reason are true and reason correctly explains assertion
A larger objective aperture collects more light (brighter image) and increases resolving power, so the reason correctly explains the assertion.
- Q6A convex lens of focal length 80 cm and a concave lens of focal length 50 cm are combined in contact. Their resulting power is:
- -6.5 D
- +6.5 D
- +7.5 D
- -0.75 D
Answer: (D) -0.75 D
P=(100)/(f₁)+(100)/(f₂)=(100)/(80)+(100)/(-50)=1.25-2.0=-0.75 D.
- Q7A coin lies at the bottom of a tank of water (n = 4/3) 16 cm deep. Its apparent depth when viewed from above is:
- 8 cm
- 16 cm
- 21.3 cm
- 12 cm
Answer: (D) 12 cm
Apparent depth = real depth / n = 16 / (4/3) = 16 × 3/4 = 12 cm.
- Q8The dispersive power of a prism material depends on:
- the size of the prism
- the angle of incidence
- only the material (its refractive indices)
- the refracting angle A only
Answer: (C) only the material (its refractive indices)
ω = (n_V – n_R)/(n_y – 1) involves only the material’s refractive indices; it is independent of the prism angle.
- Q9A microscope has an objective of focal length 2 cm, eyepiece of focal length 4 cm and the tube length of 40 cm. If the distance of distinct vision of eye is 25 cm, the magnification in the microscope is
- 150
- 125
- 250
- 100
Answer: (B) 125
D. Tests: compound microscope magnification using tube length. Why D: m=(L)/(fₒ)×(D)/(fₑ)=(40)/(2)×(25)/(4)=20× 6.25=125. Why not A: 150 needs (D)/(fₑ)=7.5, which no given data produce; it rewards rounding 6.25 up to a cleaner number [misconception A: “rounding intermediate ratios to convenient values”] Why not B: 250 comes from (40)/(2)×(25)/(2), i.e. reusing the objective focal length in the eyepiece term [misconception B: “swapping objective and eyepiece focal lengths”] Why not C: 100 follows from taking D=20 cm from memory instead of the stated 25 cm [misconception C: “memorised near point instead of given data”] Remember: microscope m≈(L)/(fₒ)·(D)/(fₑ), tube length over objective, near point over eyepiece.
- Q10A 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. A parallel beam falling on A leaves B as a parallel beam. Then the distance d (in cm) is:
- 15
- 30
- 25
- 50
Answer: (A) 15
The parallel beam converges to the focus of A (20 cm beyond A). For B to re-collimate it, this point must be at the focus of the concave lens, i.e. 5 cm before B. So d=20-5=15 cm.
- Q11An object is placed on the principal axis of a concave mirror at a distance of 1.5f (f is the focal length). The image will be at:
- 3f
- -1.5f
- 1.5f
- -3f
Answer: (D) -3f
Mirror formula (1)/(v)+(1)/(u)=(1)/(f) with u=-1.5f and f=-f: (1)/(v)=-(1)/(f)+(1)/(1.5f)=(1)/(f)(-1+(2)/(3))=-(1)/(3f), so v=-3f (real, in front of the mirror).
- Q12An 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:
- 46.0 cm
- 54.0 cm
- 37.3 cm
- 50.0 cm
Answer: (B) 54.0 cm
Objective: u=-200,f=40: (1)/(v)=(1)/(40)-(1)/(200)=(4)/(200)=(1)/(50), so v=50 cm. The eyepiece forms the final image at infinity (normal adjustment), so separation =v+fₑ=50+4=54 cm.
- Q13In a certain camera, a combination of four similar thin convex lenses are arranged axially in contact. Then the power of the combination and the total magnification in comparison to the power (p) and magnification (m) for each lens will be, respectively
- p⁴ and m⁴
- p⁴ and 4m
- 4p and 4m
- 4p and m⁴
Answer: (D) 4p and m⁴
A. Tests: combination rules for power and magnification of thin lenses in contact. Why A: For thin lenses in contact, powers add: P=p+p+p+p=4p. Each lens magnifies the image formed by the previous lens, so magnifications multiply: mₜₒₜₐₗ=m× m× m× m=m⁴. Why not B: powers add, they do not multiply [misconception B: “power compounds multiplicatively like magnification”] Why not C: magnifications multiply because each stage acts on the previous image, they do not add [misconception C: “magnification adds like power”] Why not D: this reverses both rules at once [misconception D: “swapping the two combination rules”] Remember: in contact, powers ADD, magnifications MULTIPLY.
- Q14The 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 return back through the same path after suffering reflection at the silvered surface. The refractive index μ of the prism is:
- tan A
- 2cos A
- 2sin A
- (1)/(2)cos A
Answer: (B) 2cos A
For the ray to retrace, it must strike the silvered face normally, so the refraction angle at the first face is r=A. Snell’s law: μ=(sin 2A)/(sin A)=(2sin Acos A)/(sin A)=2cos A.
- Q15An object is placed 40 cm from a concave mirror of focal length 15 cm. If the object is moved 20 cm towards the mirror, the displacement of the image will be:
- 36 cm away from the mirror
- 30 cm away from the mirror
- 30 cm towards the mirror
- 36 cm towards the mirror
Answer: (A) 36 cm away from the mirror
u₁=-40,f=-15: (1)/(v₁)=(1)/(f)-(1)/(u₁)=-(1)/(15)+(1)/(40) gives v₁=-24 cm. After moving, u₂=-20: (1)/(v₂)=-(1)/(15)+(1)/(20) gives v₂=-60 cm. Image shifts from 24 cm to 60 cm in front, i.e. 36 cm away from the mirror.
- Q16An object is placed 30 cm in front of a concave mirror of focal length 20 cm. The image distance is:
- -60 cm
- +60 cm
- -12 cm
- -30 cm
Answer: (A) -60 cm
Using 1/v + 1/u = 1/f with u = -30, f = -20: 1/v = 1/(-20) – 1/(-30) = -1/20 + 1/30 = -1/60, so v = -60 cm (real, in front).
- Q17A 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:
- μ A
- (μ A)/(2)
- (2A)/(μ)
- (A)/(2μ)
Answer: (A) μ A
B. Tests: prism geometry with normal emergence plus the small-angle form of Snell’s law. Why B: Normal emergence means the refraction angle at the second face is r₂=0. Since r₁+r₂=A, we get r₁=A. Snell’s law at the first face: sin i=μsin r₁; for a small prism the angles are small, so i≈μ r₁=μ A. Why not A: (2A)/(μ) divides by μ where entering a denser medium demands multiplying, and the 2 has no source here [misconception A: “dividing by mu when light enters glass”] Why not C: (μ A)/(2) halves r₁ as if the prism angle splits equally between the faces; that is the minimum-deviation case, but normal emergence loads the whole A onto the first face [misconception C: “r1 = A/2 in every prism problem”] Why not D: (A)/(2μ) makes both errors together [misconception D: “applying Snell’s law upside down and halving”] Remember: normal emergence forces r₂=0, r₁=A, so i=μ A for a thin prism.
- Q18The refractive index of the material of a prism is √(2) and its refracting angle is 30°. One refracting surface is silvered. 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:
- 30°
- 45°
- 60°
- 0°
Answer: (B) 45°
To retrace, the ray meets the silvered face normally, so r=A=30°. Snell’s law at entry: sin i=μsin r=√(2)sin30°=(sqrt2)/(2)=(1)/(sqrt2), so i=45°.
- Q19Pick the WRONG answer in the context with rainbow.
- The order of colours is reversed in the secondary rainbow.
- Rainbow is a combined effect of dispersion, refraction and reflection of sunlight.
- An observer can see a rainbow when his front is towards the sun.
- When the light rays undergo two internal reflections in a water drop, a secondary rainbow is formed.
Answer: (C) An observer can see a rainbow when his front is towards the sun.
A. Tests: the viewing geometry and formation mechanism of a rainbow. Why A: The question asks for the WRONG statement. A rainbow is seen only with the Sun BEHIND the observer, because raindrops send the dispersed light back towards the Sun’s side at about 40 to 42 degrees. Facing the Sun, the observer cannot receive this backscattered cone, so statement A is false and is therefore the answer. Why not B: this statement is true, a rainbow combines refraction into the drop, dispersion inside it and internal reflection at its back surface [misconception B: “rainbow is reflection alone”] Why not C: true, two internal reflections produce the secondary bow [misconception C: “the secondary bow needs a second rain shower”] Why not D: true, the colour order reverses in the secondary bow, red inside and violet outside [misconception D: “both bows share the same colour order”] Remember: back to the Sun, face the rain, then you see the bow.
- Q20In total internal reflection, when the angle of incidence is equal to the critical angle for the pair of media in contact, what will be the angle of refraction?
- equal to angle of incidence
- 90°
- 180°
- 0°
Answer: (B) 90°
B. Tests: the definition of the critical angle. Why B: The critical angle C is DEFINED as the angle of incidence in the denser medium for which the refracted ray just grazes along the interface, i.e. the angle of refraction is 90°. Snell’s law reads nsin C=1×sin 90°, which is exactly why sin C=(1)/(n). Why not A: refraction equals incidence only when the two media have the same refractive index (or at normal incidence), not at the critical angle [misconception A: “at C the ray passes straight through symmetrically”] Why not C: an angle of refraction of 180° would mean the ray runs back into the incident direction, which describes nothing physical here [misconception C: “TIR means the ray turns around, so refraction is 180 degrees”] Why not D: 0° is normal incidence with an undeviated ray, the opposite extreme [misconception D: “refraction stops, so the angle is zero”] Remember: at i=C the refracted ray skims the surface at 90°; beyond C, total internal reflection.
- Q21A ray of light travelling in a medium of refractive index μ is incident at 45° on the surface separating it from air. For which value of μ can the ray undergo total internal reflection?
- μ=1.33
- μ=1.40
- μ=1.50
- μ=1.25
Answer: (C) μ=1.50
TIR requires the angle of incidence to exceed the critical angle, i.e. sin 45°>sin i_c=(1)/(μ), so μ>√(2)≈1.414. Only μ=1.50 satisfies this.
- Q22A man 6 ft tall wants to see his entire image in a plane mirror. The minimum length of the mirror required is:
- 12 ft
- 3 ft
- 6 ft
- 2 ft
Answer: (B) 3 ft
A plane mirror needs only half the height of the object to show the full image (rays from head and feet meet at the eye after reflecting at the mid-points). Minimum length =6/2=3 ft, independent of the man’s distance from the mirror.
- Q23If the magnification produced by a mirror is +0.5, the image is:
- virtual and erect, diminished
- real and magnified
- real and inverted
- virtual and inverted
Answer: (A) virtual and erect, diminished
Positive magnification means erect/virtual; magnitude 0.5 < 1 means diminished. This corresponds to a convex mirror image.
- Q24For refraction at a single spherical surface separating media n₁ and n₂, the relation is:
- n₁/v – n₂/u = (n₁-n₂)/R
- n₂/v – n₁/u = (n₂-n₁)/R
- n₂/v + n₁/u = (n₂+n₁)/R
- 1/v – 1/u = (n₂-n₁)/R
Answer: (B) n₂/v – n₁/u = (n₂-n₁)/R
The single-surface refraction formula is n₂/v – n₁/u = (n₂ – n₁)/R, with v in the second medium and u in the first.
- Q25A ray is incident at an angle of incidence i on one surface of a small-angle prism (angle A) and emerges normally from the opposite surface. If the refractive index of the prism material is μ, then the angle of incidence i is nearly equal to:
- μ A
- (μ A)/(2)
- (A)/(2μ)
- (A)/(μ)
Answer: (A) μ A
Emergence normal to the second face means r₂=0, so r₁=A. For a small angle Snell’s law gives i≈μ r₁=μ A.