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NEET Exam  >  NEET Test  >  Physics Class 12  >  31 Year NEET Previous Year Questions: Wave Optics - NEET MCQ

31 Year NEET Previous Questions: Wave Optics Free MCQ Practice Test [Updated]


MCQ Practice Test & Solutions: 31 Year NEET Previous Year Questions: Wave Optics (21 Questions)

You can prepare effectively for NEET Physics Class 12 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "31 Year NEET Previous Year Questions: Wave Optics". These 21 questions have been designed by the experts with the latest curriculum of NEET 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 40 minutes
  • - Number of Questions: 21

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31 Year NEET Previous Year Questions: Wave Optics - Question 1
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Which one of the following phenomena is notexplained by Huygen’s construction ofwavefront? [1988]

Detailed Solution: Question 1

Huygen's construction of wavefront does not apply to origin of spectra which is explained by quantum theory

31 Year NEET Previous Year Questions: Wave Optics - Question 2
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Which of the following phenomenon is notcommon to sound and light waves ?      [1988]

Detailed Solution: Question 2

Sound waves can not be polarised as they are longitudinal. Light waves can be polarised as they are transverse.

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31 Year NEET Previous Year Questions: Wave Optics - Question 3
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Interference is possible in [1989]

Detailed Solution: Question 3

Interference is a wave phenomenon shown by both the light waves and sound waves.

31 Year NEET Previous Year Questions: Wave Optics - Question 4
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The Young’s double slit experiment is performed with blue and with green light of wavelengths 4360Å and 5460Å respectively. If x is the distance of 4th maxima from the central one, then           [1990]

Detailed Solution: Question 4

Distance of nth maxima, 

31 Year NEET Previous Year Questions: Wave Optics - Question 5
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In Young’s double slit experiment, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed in water of refrative index  4/3, without disturbing the geometrical arrangement, the new fringe width will be [1990]

Detailed Solution: Question 5

31 Year NEET Previous Year Questions: Wave Optics - Question 6
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In Young’s experiment, two coherent sources areplaced 0.90 mm apart and fringe are observedone metre away. If it produces second dark fringeat a distance of 1 mm from central fringe, thewavelength of monochromatic light used wouldbe               [1991, 1992]

Detailed Solution: Question 6

For dark fringe

λ = 0.6x10-6 m = 6 × 10–5 cm

31 Year NEET Previous Year Questions: Wave Optics - Question 7
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Ratio of intensities of two waves are given by 4 : 1. Then the ratio of the amplitudes of the twowaves is   [1991]

Detailed Solution: Question 7

31 Year NEET Previous Year Questions: Wave Optics - Question 8
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In Young’s double slit experiment carried out withlight of wavelength (λ) = 5000Å, the distancebetween the slits is 0.2 mm and the screen is at 200 cm from the slits. The central maximum is at x = 0. The third maximum (taking the central maximum as zero th maximum) will be at x equal to           [1992]

Detailed Solution: Question 8

31 Year NEET Previous Year Questions: Wave Optics - Question 9
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If yellow light emitted by sodium lamp in Young’s double slit experiment is replaced by a monochromatic blue light of the same intensity        [1992]

Detailed Solution: Question 9

  1. Fringe Width Formula:
    In Young’s double slit experiment, the fringe width (beta) is given by:
    beta = (lambda * D) / d
    where:

    • lambda = wavelength of light

    • D = distance between the slits and the screen

    • d = distance between the two slits

  2. Compare Wavelengths:

    • Yellow light (from a sodium lamp) has a wavelength of about 589 nm.

    • Blue light has a shorter wavelength, typically around 450 to 495 nm (for example, 475 nm).

    • So, blue light has a shorter wavelength than yellow light.

    • Since D and d stay the same (setup doesn't change), the fringe width depends directly on the wavelength (lambda).

  3. Effect on Fringe Width:

    • Fringe width (beta) is directly proportional to lambda.

    • Since blue light has a shorter wavelength, the fringe width will decrease when we replace yellow light with blue light.

  4. Intensity Note:

    • The question says the intensity is the same. Intensity affects brightness, but not the fringe width. Option d) talks about intensity, which is unrelated here.

a) Fringe width will decrease

  • a) Correct: Fringe width decreases because blue light has a shorter wavelength than yellow light.

  • b) Incorrect: Fringe width would increase if the wavelength increased (like with red light).

  • c) Incorrect: Fringe width changes because the wavelength is changing.

  • d) Incorrect: Intensity being the same doesn’t affect fringe width, and the question asks about fringe width, not brightness.

Thus, the correct answer is a).

31 Year NEET Previous Year Questions: Wave Optics - Question 10
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Interference was observed in interferencechamber where air was present, now the chamberis evacuated, and if the same light is used, acareful observer will see           [1993]

Detailed Solution: Question 10

In vaccum, λ increases very slightly compared to that in air. As β∝ λ , therefore, width of interference fringe increases slightly.

31 Year NEET Previous Year Questions: Wave Optics - Question 11
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A parallel beam of monochromatic light ofwavelength 5000Å is incident normally on asingle narrow slit of width 0.001 mm. The light isfocussed by a convex lens on a screen placed infocal plane. The first minimum will be formed forthe angle of diffraction equal to        [1993]

Detailed Solution: Question 11

For first minimum, a sin θ = nλ = 1λ

31 Year NEET Previous Year Questions: Wave Optics - Question 12
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In a Fresnel biprism experiment, the two positionsof lens give separation between the slits as 16cm and 9 cm respectively. What is the actualdistance of separation?    [1996]

Detailed Solution: Question 12

Separation between slits are (r1=) 16 cm and (r2=) 9 cm.
Actual distance of separation

31 Year NEET Previous Year Questions: Wave Optics - Question 13
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Colours appear on a thin soap film and on soapbubbles due to the phenomenon of [1999]

Detailed Solution: Question 13

We know that the colours for which the
condition of constructive interference is
satisfied are observed in a given region of the film. The path difference between the light waves reaching the eye changes when the position of the eye is changed. Therefore, colours appear on a thin soap film or soap bubbles due to the phenomenon of interference.

31 Year NEET Previous Year Questions: Wave Optics - Question 14
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Which of the following shows particle nature oflight?      [2001]

Detailed Solution: Question 14

Photoelectric effect shows particle nature of light as light consists of packets of energy called photons or quanta

31 Year NEET Previous Year Questions: Wave Optics - Question 15
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A paper, with two marks having separation d, isheld normal to the line of sight of an observer ata distance of 50m. The diameter of the eye-lensof the observer is 2 mm. Which of the followingis the least value of d, so that the marks can beseen as separate ? The mean wavelength ofvisible light may be taken as 5000 Å.     [2002]

Detailed Solution: Question 15

31 Year NEET Previous Year Questions: Wave Optics - Question 16
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The periodic waves of intensities I1 and I2 pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is:

Detailed Solution: Question 16

We know that the resultant of two intensities with phase difference of angle a can be represented as, I = I1+ I2 + 2( I1.I2 )1/2. cos a
Thus we get Imax =  I1+ I2 + 2( I1.I2 )½
And Imin =  I1+ I2 - 2( I1.I2 )½
Thus we get the sum =  2( I1+ I2 )

31 Year NEET Previous Year Questions: Wave Optics - Question 17
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A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter
2/d  in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be
respectively:

Detailed Solution: Question 17

By covering aperture, focal length does not change. But intensity is reduced by 1/4 times, as aperture diameter d/2 is covered.

∴ New focal length = f and intensity = 3I/4

31 Year NEET Previous Year Questions: Wave Optics - Question 18
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In Young’s double slit experiment, the slits are 2 mmapart and are illuminated by photons of twowavelengths λ1 = 12000Å and λ2 = 10000Å. Atwhat minimum distance from the common centralbright fringe on the screen 2 m from the slit will abright fringe from one interference patterncoincide with a bright fringe from the other ?    [NEET 2013]

Detailed Solution: Question 18

∴ n1 λ1 = n2λ2
⇒ n1 × 12000 × 10–10 = n2 × 10000 × 10–10
or, n (12000 × 10–10) = (n + 1) (10000 × 10–10)
⇒n = 5

(∵ λ1 = 1200 x 10-10m;  λ2 = 10000 x 10-10m)

(∵ d = 2 mm and D = 2m)
= 5 × 12 × 10–4 m
= 60 × 10–4 m
= 6 × 10–3m = 6 mm

31 Year NEET Previous Year Questions: Wave Optics - Question 19
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A parallel beam of fast moving electrons isincident normally on a narrow slit. A fluorescentscreen is placed at a large distance from the slit.If the speed of the electrons is increased, whichof the following statements is correct ?              [NEET 2013]

Detailed Solution: Question 19

Angular width, 

[For central maxima n = 1]
Hence, with increase in speed of
electrons angular width of central
maximum decreases.

31 Year NEET Previous Year Questions: Wave Optics - Question 20
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In Young’s double slit experiment the distancebetween the slits and the screen is doubled. Theseparation between the slits is reduced to half.As a result the fringe width           [NEET Kar. 2013]

Detailed Solution: Question 20

Fringe width  

From question D' = 2D and d' = d/2

31 Year NEET Previous Year Questions: Wave Optics - Question 21
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A parallel beam of light of wavelength λ is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is       [NEET Kar. 2013]

Detailed Solution: Question 21

Conditions for diffraction minima are
Path diff. Δx = nλ and Phase diff. δφ = 2nπ
Path diff. = nλ = 2λ
Phase diff. = 2nπ = 4π (∵n = 2)

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This test contains 21 MCQs on 31 Year NEET Previous Year Questions: Wave Optics with detailed solutions for each question. It is part of the Physics Class 12 course on EduRev, designed to align with the current NEET syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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