Wave Optics JEE Main previous year questions with solutions

5 solved JEE Main questions on Wave Optics, free to read — no sign-in needed. The full chapter has 188 questions; sign in to attempt the remaining 183 in the exam simulator.

1. Q1JEE Main 2026 (08 Apr, Shift 2)Diffraction of Light
   Some distant star is to be observed by some telescope of diameter of objective lens $a$, at an angular resolution of $3.0\times 10^{-7}$ radian. If the wavelength of light from the star reaching the telescope is $500$ nm, the minimum diameter of the objective lens of the telescope is ________ cm. (nearest integer)
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   Answer: 203

   The angular resolution $\Delta \theta$ of a telescope is given by the formula: $\Delta \theta = \dfrac{1.22 \lambda}{a}$ where $\lambda$ is the wavelength of light and $a$ is the diameter of the objective lens. Given: $\Delta \theta = 3.0 \times 10^{-7}$ rad $\lambda = 500 \text{ nm} = 500 \times 10^{-9}$ m $= 5 \times 10^{-7}$ m Substituting the values into the formula: $3.0 \times 10^{-7} = \dfrac{1.22 \times 5 \times 10^{-7}}{a}$ $a = \dfrac{1.22 \times 5 \times 10^{-7}}{3.0 \times 10^{-7}}$ $a = \dfrac{6.1}{3.0}$ m $a = 2.0333...$ m Converting the diameter into centimeters: $a = 2.0333... \times 100$ cm $= 203.33$ cm Rounding to the nearest integer, we get $203$ cm. Answer: $203$
2. Q2JEE Main 2025 (03 Apr, Shift 1)Interference of Light
   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 ______.
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   Answer: 24

   $\begin{aligned} & \mathrm{I}_{\max }=\left(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}\right)^2 \\ & =(\sqrt{4 \mathrm{I}}+\sqrt{9 \mathrm{I}})^2=25 \mathrm{I} \\ & \mathrm{I}_{\min }=\left(\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}\right)^2 \\ & =(\sqrt{4 \mathrm{I}}-\sqrt{9 \mathrm{I}})^2=\mathrm{I} \\ & \mathrm{I}_{\max }-\mathrm{I}_{\min }=24 \mathrm{I} \\ & \mathrm{x}=24\end{aligned}$
3. Q3JEE Main 2024 (30 Jan, Shift 2)Polarization of Light
   A beam of unpolarised light of intensity ${I}_{0}$ is passed through a polaroid $A$ and then through another polaroid $B$ which is oriented so that its principal plane makes an angle of $45^{\circ}$ relative to that of $A$. The intensity of emergent light is :
   1. A.$\dfrac{{I}_{0}}{4}$
   2. B.${I}_{0}$
   3. C.$\dfrac{{I}_{0}}{2}$
   4. D.$\dfrac{{I}_{0}}{8}$
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   Answer: (A)

   Intensity of polarised light from polaroid $A$ will be $\dfrac{{I}_{0}}{2}$. Therefore, Intensity of emergent light $=\dfrac{{I}_{0}}{2}{\cos }^{2}45^{\circ}=\dfrac{{I}_{0}}{4}$
4. Q4JEE Main 2023 (15 Apr, Shift 1)Youngs Double Slit Experiment
   A single slit of width $a$ is illuminated by a monochromatic light of wavelength $600nm$. The value of $a$ for which first minimum appears at $\theta ={30}^{o}$ on the screen will be :
   1. A.$1.2\mu m$
   2. B.$3\mu m$
   3. C.$1.8\mu m$
   4. D.$0.6\mu m$
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   Answer: (A)

   For the first minima $a\sin \theta =\lambda$ The given data is $\theta =30^{\circ} \lambda =600nm$ The value of $a$ is $a=\dfrac{\lambda }{\sin 30^{\circ}}=\dfrac{600\times {10}^{-9}}{\dfrac{1}{2}}=1.2\mu m$
5. Q5JEE Main 2026 (04 Apr, Shift 1)Diffraction of Light
   A telescope with objective diameter $R$ is used to observe a distant star emitting light of wavelength $500$ nm, at a resolution of $5 \times 10^{-7}$ radian. The value of $R$ is _____ cm.
   1. A.$61$
   2. B.$122$
   3. C.$244$
   4. D.$305$
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   Answer: (B)

   The limit of resolution of a telescope is given by the formula: $\Delta \theta = \dfrac{1.22 \lambda}{R}$ where $\lambda$ is the wavelength of light and $R$ is the diameter of the objective lens. Given: $\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} = 5 \times 10^{-7} \text{ m}$ $\Delta \theta = 5 \times 10^{-7} \text{ rad}$ Substituting the values into the formula: $5 \times 10^{-7} = \dfrac{1.22 \times 5 \times 10^{-7}}{R}$ $R = 1.22 \text{ m}$ Converting to centimeters: $R = 1.22 \times 100 \text{ cm} = 122 \text{ cm}$ Answer: $122$