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📘 Wave Optics — JEE Mains PYQs
Step-by-Step Solutions with Detailed Explanations
Master Wave Optics for JEE Main with previous year questions on interference, diffraction, and polarization. Each solution is explained in a simple, step-by-step manner to boost your score.
Wave Optics — JEE Mains PYQs
JEE Main 2018–2025Q1. JEE Main 2025 (8 April Shift 2)
In a Young's double slit experiment, the source is white light. One of the slits is covered by red filter and another by a green filter. In this case :
(1) There shall be an interference pattern for red distinct from that for green.
(2) There shall be no interference fringes.
(3) There shall be alternate interference fringes of red and green.
(4) There shall be an interference pattern, where each fringe's pattern center is green and outer edges is red.
(1) There shall be an interference pattern for red distinct from that for green.
(2) There shall be no interference fringes.
(3) There shall be alternate interference fringes of red and green.
(4) There shall be an interference pattern, where each fringe's pattern center is green and outer edges is red.
✅ Answer & Explanation:
Correct Option: (2)
Interference occurs only between waves from coherent sources having the same frequency and a constant phase difference. Red and green light have different frequencies; waves of different frequencies are incoherent and cannot produce a stable interference pattern. Thus, no fringes will be observed.
Interference occurs only between waves from coherent sources having the same frequency and a constant phase difference. Red and green light have different frequencies; waves of different frequencies are incoherent and cannot produce a stable interference pattern. Thus, no fringes will be observed.
Q2. JEE Main 2025 (4 April Shift 2)
In a Young's double slit experiment, two slits are located 1.5 mm apart. The distance of screen from slits is 2 m and the wavelength of the source is 400 nm. If the 20 maxima of the double slit pattern are contained within the centre maximum of the single slit diffraction pattern, then the width of each slit is \(x \times 10^{-3} \, \text{cm}\), where \(x\)-value is ________.
✅ Answer & Explanation:
Correct Answer: 15
- Width of central maximum in diffraction: \(\theta = \frac{2\lambda}{a}\) (where \(a\) is slit width).
- Angular width of one interference fringe: \(\Delta\theta = \frac{\lambda}{d}\) (where \(d\) is slit separation).
- Given 20 interference maxima fit in the central diffraction maximum: \(20 \times \frac{\lambda}{d} = \frac{2\lambda}{a}\).
- \(a = \frac{2d}{20} = \frac{d}{10} = \frac{1.5 \, \text{mm}}{10} = 0.15 \, \text{mm}\).
- \(a = 0.015 \, \text{cm} = 15 \times 10^{-3} \, \text{cm}\). Thus, \(x = 15\).
Q3. JEE Main 2025 (3 April Shift 2)
Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is :
(1) \((2\sqrt{2} + 1) : (2\sqrt{2} - 1)\) (2) \((3 + 2\sqrt{2}) : (3 - 2\sqrt{2})\)
(3) 9 : 1 (4) 3 : 1
(1) \((2\sqrt{2} + 1) : (2\sqrt{2} - 1)\) (2) \((3 + 2\sqrt{2}) : (3 - 2\sqrt{2})\)
(3) 9 : 1 (4) 3 : 1
✅ Answer & Explanation:
Correct Option: (2)
- Intensity \(I \propto \text{Slit Width}\). Let \(I_1 = I\) and \(I_2 = 2I\).
- \(\frac{I_{max}}{I_{min}} = \left( \frac{\sqrt{I_2} + \sqrt{I_1}}{\sqrt{I_2} - \sqrt{I_1}} \right)^2 = \left( \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right)^2\).
- Ratio \(= \frac{2 + 1 + 2\sqrt{2}}{2 + 1 - 2\sqrt{2}} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}\).
Q4. JEE Main 2025 (3 April Shift 1)
Two coherent monochromatic light beams of intensities 4I and 9I are superimposed. The difference between the maximum and minimum intensities in the resulting interference pattern is xI. The value of x is ______.
✅ Answer & Explanation:
Correct Answer: 24
- \(I_{max} = (\sqrt{4I} + \sqrt{9I})^2 = (2\sqrt{I} + 3\sqrt{I})^2 = 25I\).
- \(I_{min} = (\sqrt{9I} - \sqrt{4I})^2 = (3\sqrt{I} - 2\sqrt{I})^2 = I\).
- Difference \(= 25I - I = 24I\). Thus, \(x = 24\).
Q5. JEE Main 2025 (28 Jan Shift 1)
A double slit interference experiment performed with a light of wavelength 600 nm forms an interference fringe pattern on a screen with 10th bright fringe having its centre at a distance of 10 mm from the central maximum. Distance of the centre of the same 10th bright fringe from the central maximum when the source of light is replaced by another source of wavelength 660 nm would be _________ mm.
✅ Answer & Explanation:
Correct Answer: 11
- Position of \(n\)-th bright fringe \(y_n = \frac{n \lambda D}{d}\).
- Since \(n, D, d\) are constant, \(y \propto \lambda\).
- \(\frac{y_2}{y_1} = \frac{\lambda_2}{\lambda_1} \implies \frac{y_2}{10} = \frac{660}{600} = 1.1\).
- \(y_2 = 1.1 \times 10 = \mathbf{11 \, \text{mm}}\).
Q6. JEE Main 2025 (24 Jan Shift 1)
The Young's double slit interference experiment is performed using light consisting of 480 nm and 600 nm wavelengths to form interference patterns. The least number of the bright fringes of 480 nm light that are required for the first coincidence with the bright fringes formed by 600 nm light is :
(1) 5 (2) 4
(3) 6 (4) 8
(1) 5 (2) 4
(3) 6 (4) 8
✅ Answer & Explanation:
Correct Option: (1)
- For coincidence, \(n_1 \lambda_1 = n_2 \lambda_2\).
- \(n_1(480) = n_2(600) \implies \frac{n_1}{n_2} = \frac{600}{480} = \frac{5}{4}\).
- The least integer value for \(n_1\) (fringe of 480 nm light) is 5.
Q7. JEE Main 2025 (22 Jan Shift 1)
Assertion (A): If Young's double slit experiment is performed in an optically denser medium than air, then the consecutive fringes come closer.
Reason (R): The speed of light reduces in an optically denser medium than air while its frequency does not change.
(1) Both (A) and (R) are true but (R) is not the correct explanation of (A)
(2) Both (A) and (R) are true and (R) is the correct explanation of (A)
(3) (A) is true but (R) is false
(4) (A) is false but (R) is true
Reason (R): The speed of light reduces in an optically denser medium than air while its frequency does not change.
(1) Both (A) and (R) are true but (R) is not the correct explanation of (A)
(2) Both (A) and (R) are true and (R) is the correct explanation of (A)
(3) (A) is true but (R) is false
(4) (A) is false but (R) is true
✅ Answer & Explanation:
Correct Option: (2)
Fringe width \(\beta = \frac{\lambda D}{d}\). In a denser medium, the wavelength \(\lambda\) decreases (\(\lambda' = \lambda/n\)) because the speed reduces while frequency remains constant. Since \(\lambda\) decreases, the fringe width \(\beta\) also decreases, causing fringes to come closer.
Fringe width \(\beta = \frac{\lambda D}{d}\). In a denser medium, the wavelength \(\lambda\) decreases (\(\lambda' = \lambda/n\)) because the speed reduces while frequency remains constant. Since \(\lambda\) decreases, the fringe width \(\beta\) also decreases, causing fringes to come closer.
Q8. JEE Main 2023 (06 Apr Shift 2)
A beam of light consisting of two wavelengths \(7000 \, \unicode{x212B}\) and \(5500 \, \unicode{x212B}\) is used to obtain interference pattern in Young's double slit experiment. The distance between the slits is 2.5 mm and the distance between the plane of slits and the screen is 150 cm. The least distance from the central fringe, where the bright fringes due to both the wavelengths coincide, is \(n \times 10^{-5} \, \text{m}\). The value of \(n\) is _____.
✅ Answer & Explanation:
Correct Answer: 462
- Coincidence condition: \(n_1 \lambda_1 = n_2 \lambda_2 \implies n_1(7000) = n_2(5500) \implies \frac{n_1}{n_2} = \frac{11}{14}\).
- Smallest \(n_1 = 11\). Distance \(y = \frac{n_1 \lambda_1 D}{d}\).
- \(y = \frac{11 \times 700 \times 10^{-9} \times 1.5}{2.5 \times 10^{-3}} = \frac{11550 \times 10^{-9}}{2.5 \times 10^{-3}} = 4.62 \times 10^{-3} \, \text{m} = 462 \times 10^{-5} \, \text{m}\).
- Thus, \(n = 462\).
Q9. JEE Main 2022 (27 Jul Shift 2)
Two coherent sources of light interfere. The intensity ratio of two sources is 1 : 4. For this interference pattern if the value of \(\frac{I_{max} + I_{min}}{I_{max} - I_{min}}\) is equal to \(\frac{2\alpha+1}{\beta+3}\), then \(\alpha/\beta\) will be :
(1) 1.5 (2) 2
(3) 0.5 (4) 1
(1) 1.5 (2) 2
(3) 0.5 (4) 1
✅ Answer & Explanation:
Correct Option: (2)
- \(I_{max} : I_{min} = (\sqrt{4} + \sqrt{1})^2 : (\sqrt{4} - \sqrt{1})^2 = 9 : 1\).
- \(\frac{I_{max} + I_{min}}{I_{max} - I_{min}} = \frac{10}{8} = 1.25\).
- From answer key, \(\alpha=2\) and \(\beta=1\)? Note: Re-check algebraic matching in source. Key specifies option (2).
Q10. JEE Main 2021 (25 Jul Shift 1)
In the Young's double slit experiment, the distance between the slits varies in time as \(d(t) = d_0 + a_0 \sin\omega t\); where \(d_0, \omega\) and \(a_0\) are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as:
(1) \(\frac{2\lambda D d_0}{(d_0^2 - a_0^2)}\) (2) \(\frac{2\lambda D a_0}{(d_0^2 - a_0^2)}\)
(3) \(\frac{\lambda D}{d_0^2} a_0\) (4) \(\frac{\lambda D}{d_0 + a_0}\)
(1) \(\frac{2\lambda D d_0}{(d_0^2 - a_0^2)}\) (2) \(\frac{2\lambda D a_0}{(d_0^2 - a_0^2)}\)
(3) \(\frac{\lambda D}{d_0^2} a_0\) (4) \(\frac{\lambda D}{d_0 + a_0}\)
✅ Answer & Explanation:
Correct Option: (2)
- Fringe width \(\beta = \frac{\lambda D}{d}\).
- \(\beta_{max} = \frac{\lambda D}{d_0 - a_0}\) and \(\beta_{min} = \frac{\lambda D}{d_0 + a_0}\).
- Difference \(\Delta\beta = \lambda D \left( \frac{1}{d_0 - a_0} - \frac{1}{d_0 + a_0} \right) = \lambda D \frac{(d_0 + a_0) - (d_0 - a_0)}{d_0^2 - a_0^2} = \frac{2\lambda D a_0}{d_0^2 - a_0^2}\).
Q11. JEE Main 2025 (3 April Shift 2)
A monochromatic light of frequency \(5 \times 10^{14} \, \text{Hz}\) travelling through air, is incident on a medium of refractive index '2'. Wavelength of the refracted light will be :
(1) 300 nm (2) 600 nm
(3) 400 nm (4) 500 nm
(1) 300 nm (2) 600 nm
(3) 400 nm (4) 500 nm
✅ Answer & Explanation:
Correct Option: (1)
- Wavelength in air \(\lambda_{air} = \frac{c}{f} = \frac{3 \times 10^8}{5 \times 10^{14}} = 6 \times 10^{-7} \, \text{m} = 600 \, \text{nm}\).
- Wavelength in medium \(\lambda_{med} = \frac{\lambda_{air}}{n} = \frac{600}{2} = \mathbf{300 \, \text{nm}}\).
Q12. JEE Main 2025 (28 Jan Shift 2)
Which of the following phenomena can not be explained by wave theory of light?
(1) Compton effect (2) Refraction of light
(3) Reflection of light (4) Diffraction of light
(1) Compton effect (2) Refraction of light
(3) Reflection of light (4) Diffraction of light
✅ Answer & Explanation:
Correct Option: (1)
The Compton effect involves the scattering of a photon by a charged particle (usually an electron), which results in a decrease in energy (increase in wavelength) of the photon. This phenomenon proves the particle nature of light and cannot be explained using classical wave theory.
The Compton effect involves the scattering of a photon by a charged particle (usually an electron), which results in a decrease in energy (increase in wavelength) of the photon. This phenomenon proves the particle nature of light and cannot be explained using classical wave theory.
Q13. JEE Main 2023 (01 Feb Shift 1)
'n' polarizing sheets are arranged such that each makes an angle \(45^\circ\) with the proceeding sheet. An unpolarized light of intensity \(I\) is incident into this arrangement. The output intensity is found to be \(I/64\). The value of \(n\) will be:
(1) 3 (2) 6
(3) 5 (4) 4
(1) 3 (2) 6
(3) 5 (4) 4
✅ Answer & Explanation:
Correct Option: (2)
- Intensity after first polarizer: \(I_1 = \frac{I}{2}\).
- Each subsequent sheet (\(n-1\) sheets) reduces intensity by factor of \(\cos^2 45^\circ = \frac{1}{2}\).
- Final intensity \(I_{out} = \frac{I}{2} \times \left( \frac{1}{2} \right)^{n-1} = \frac{I}{2^n}\).
- \(\frac{I}{64} = I/2^n \implies 2^n = 64 \implies \mathbf{n = 6}\).
Q14. JEE Main 2022 (24 Jun Shift 1)
Sodium light of wavelengths 650 nm and 655 nm is used to study diffraction at a single slit of aperture 0.5 mm. The distance between the slit and the screen is 2.0 m. The separation between the positions of the first maxima of diffraction pattern obtained in the two cases is _____ \(\times 10^{-5} \, \text{m}\).
✅ Answer & Explanation:
Correct Answer: 3
- Position of first diffraction maximum: \(y \approx \frac{3\lambda D}{2a}\).
- Separation \(\Delta y = \frac{3D}{2a} (\lambda_2 - \lambda_1) = \frac{3 \times 2}{2 \times 0.5 \times 10^{-3}} (5 \times 10^{-9})\).
- \(\Delta y = \frac{6}{10^{-3}} \times 5 \times 10^{-9} = 30 \times 10^{-6} = 3 \times 10^{-5} \, \text{m}\).
- Value is 3.
Q15. JEE Main 2020 (09 Jan Shift 1)
The aperture diameter of a telescope is 5 m. The separation between the moon and the earth is \(4 \times 10^5 \, \text{km}\). With light of wavelength of \(5500 \, \unicode{x212B}\), the minimum separation between objects on the surface of moon, so that they are just resolved, is close to:
(1) 60 m (2) 20 m
(3) 200 m (4) 600 m
(1) 60 m (2) 20 m
(3) 200 m (4) 600 m
✅ Answer & Explanation:
Correct Option: (1)
- Angular resolution \(\Delta\theta = \frac{1.22\lambda}{D}\).
- \(\Delta\theta = \frac{1.22 \times 5500 \times 10^{-10}}{5} = 1.342 \times 10^{-7} \, \text{rad}\).
- Min separation \(x = R \Delta\theta = (4 \times 10^8) \times (1.342 \times 10^{-7}) \approx 53.68 \, \text{m}\).
- This value is close to 60 m.
Q16. JEE Main 2020 (04 Sep Shift 1)
A beam of plane polarized light of large cross-sectional area and uniform intensity of \(3.3 \, \text{W m}^{-2}\) falls normally on a polarizer (cross-sectional area \(3 \times 10^{-4} \, \text{m}^2\)), which rotates about its axis with an angular speed of \(31.4 \, \text{rad s}^{-1}\). The energy of light passing through the polarizer per revolution, is close to:
(1) \(1.0 \times 10^{-5} \, \text{J}\) (2) \(1.0 \times 10^{-4} \, \text{J}\)
(3) \(1.5 \times 10^{-4} \, \text{J}\) (4) \(5.0 \times 10^{-4} \, \text{J}\)
(1) \(1.0 \times 10^{-5} \, \text{J}\) (2) \(1.0 \times 10^{-4} \, \text{J}\)
(3) \(1.5 \times 10^{-4} \, \text{J}\) (4) \(5.0 \times 10^{-4} \, \text{J}\)
✅ Answer & Explanation:
Correct Option: (2)
- Average intensity transmitted through a rotating polarizer: \(I_{avg} = I_0/2\).
- Average power \(P_{avg} = I_{avg} \times A = (3.3 / 2) \times (3 \times 10^{-4}) = 4.95 \times 10^{-4} \, \text{W}\).
- Time for one revolution \(T = 2\pi/\omega = (2 \times 3.14) / 31.4 = 0.2 \, \text{s}\).
- Energy \(E = P_{avg} \times T = (4.95 \times 10^{-4}) \times 0.2 \approx 0.99 \times 10^{-4} \, \text{J}\).
- This is close to \(1.0 \times 10^{-4} \, \text{J}\).
Q17. JEE Main 2025 (24 Jan Shift 2)
In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of \(P_1\) and \(P_2\) are orthogonal to each other. The polarizer \(P_3\) covers both the slits with its transmission axis at \(45^\circ\) to those of \(P_1\) and \(P_2\). An unpolarized light of wavelength \(\lambda\) and intensity \(I_0\) is incident on \(P_1\) and \(P_2\). The intensity at a point after \(P_3\) where the path difference between the light waves from \(s_1\) and \(s_2\) is \(\lambda/3\), is :
(1) \(I_0/2\) (2) \(I_0/4\)
(3) \(I_0/3\) (4) \(I_0\)
(3) \(I_0/3\) (4) \(I_0\)
✅ Answer & Explanation:
Correct Option: (2)
Light from \(s_1\) and \(s_2\) becomes plane polarized orthogonally. Polarizer \(P_3\) allows components of both waves to interfere by resolving them along its \(45^\circ\) axis. With a path difference of \(\lambda/3\) (phase difference \(2\pi/3\)), the vector summation of components results in a net intensity of \(I_0/4\).
Light from \(s_1\) and \(s_2\) becomes plane polarized orthogonally. Polarizer \(P_3\) allows components of both waves to interfere by resolving them along its \(45^\circ\) axis. With a path difference of \(\lambda/3\) (phase difference \(2\pi/3\)), the vector summation of components results in a net intensity of \(I_0/4\).
Q18. JEE Main 2020 (06 Sep Shift 1)
In the figure below, P and Q are two equally intense coherent sources emitting radiation of wavelength 20 m. The separation between P and Q is 5 m and the phase of P is ahead of that of Q by \(90^\circ\). A, B and C are three distinct point of observation, each equidistant from the midpoint of PQ. The intensities of radiation at A, B, C will be in the ratio :
(1) 0 : 1 : 4 (2) 2 : 1 : 0
(3) 0 : 1 : 2 (4) 4 : 1 : 0
(3) 0 : 1 : 2 (4) 4 : 1 : 0
✅ Answer & Explanation:
Correct Option: (2)
Calculating the net phase difference (path difference + initial phase) at each point reveals constructive interference at A (\(2I\)), partial at B (\(I\)), and destructive at C (0), resulting in the ratio 2 : 1 : 0.
Calculating the net phase difference (path difference + initial phase) at each point reveals constructive interference at A (\(2I\)), partial at B (\(I\)), and destructive at C (0), resulting in the ratio 2 : 1 : 0.
📘 Exam Preparation Tip for Wave Optics
- Young's Double Slit Experiment (YDSE): Fringe width \(\beta = \frac{\lambda D}{d}\). For bright fringes: \(y_n = \frac{n\lambda D}{d}\). For dark fringes: \(y_n = \frac{(2n-1)\lambda D}{2d}\).
- Intensity in Interference: \(I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos\phi\). \(I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2\) and \(I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2\).
- Diffraction at Single Slit: Minima at \(a\sin\theta = n\lambda\). The central maximum is twice as wide as other maxima.
- Resolving Power: For a telescope, the angular resolution \(\Delta\theta = \frac{1.22\lambda}{D}\). For a microscope, resolving power is inversely proportional to the wavelength used.
- Polarization: Malus's Law: \(I = I_0 \cos^2\theta\). Unpolarized light intensity halves when passing through a polarizer.
- Coherence: For stable interference, sources must be coherent (same frequency and constant phase difference). Different frequencies cannot produce a stable interference pattern.