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 191 questions; sign in to attempt the remaining 186 in the exam simulator.

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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 (07 Apr, Shift 1)Electromagnetic Waves and Maxwell's equations
   Two plane polarized light waves combine at a certain point whose electric field components are $\begin{aligned} & \mathrm{E}_1=\mathrm{E}_0 \sin \omega \mathrm{t} \\ & \mathrm{E}_2=\mathrm{E}_0 \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right)\end{aligned}$ Find the amplitude of the resultant wave.
   1. A.0.9 E
   2. B.$\mathrm{E}_0$
   3. C.$1.7 \mathrm{E}_0$
   4. D.$3.4 \mathrm{E}_0$
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   Answer: (C)

   $\mathrm{E}=\sqrt{\left(\mathrm{E}_0\right)^2+\left(\mathrm{E}_0\right)^2+2\left(\mathrm{E}_0\right)\left(\mathrm{E}_0\right) \cos \frac{\pi}{3}}$ $\mathrm{E}=\sqrt{2 \mathrm{E}_0^2+\mathrm{E}_0^2}=\sqrt{3} \mathrm{E}_0=1.73 \mathrm{E}_0$
3. Q3JEE Main 2024 (06 Apr, Shift 2)Interference of Light
   Two coherent monochromatic light beams of intensities I and $4 \mathrm{I}$ are superimposed. The difference between maximum and minimum possible intensities in the resulting beam is $x \mathrm{I}$. The value of $x$ is _________.
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   Answer: 8

   $\begin{aligned} & I_{\max }=(\sqrt{\mathrm{I}}+\sqrt{4 \mathrm{I}})^2=9 \mathrm{I} \\ & \mathrm{I}_{\min }=(\sqrt{4 \mathrm{I}}-\sqrt{\mathrm{I}})^2=\mathrm{I} \\ & \therefore \mathrm{I}_{\max }-\mathrm{I}_{\min }=8 \mathrm{I}\end{aligned}$
4. Q4JEE Main 2023 (01 Feb, Shift 1)Polarization of Light
   ‘$n$’ 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 $\dfrac{I}{64}$ . The value of $n$ will be:
   1. A.$3$
   2. B.$6$
   3. C.$5$
   4. D.$4$
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   Answer: (B)

   When an unpolarized light passes through a polariser, the intensity of the emergent light from the polariser depends on the square of the cosine of the angle between the plane of the polariser and the direction of the incident light. The formula to calculate the intensity $\left({I}_{1}\right)$ of the emergent light from a polariser which makes an angle $\theta$ with the direction of the incident light is given by ${I}_{1}=I{\cos }^{2}\theta ..................(1)$ where, $I$ is the intensity of the incident light. Substitute $45^{\circ}$ for $\theta$ into equation (1) to obtain the intensity of the first emergent light. $\begin{matrix}{I}_{1} & = & I{\cos }^{2}45^{\circ} \\ & = & \dfrac{I}{2}\end{matrix}$ In a similar manner, the intensity $\left({I}_{2}\right)$ of the second emergent light can be written as $\begin{matrix}{I}_{2} & = & {I}_{1}{\cos }^{2}45^{\circ} \\ & = & \dfrac{{I}_{1}}{2} \\ & = & \dfrac{I}{4} \\ & = & \dfrac{I}{{2}^{2}}\end{matrix}$ Thus, for $n$ number of such polarisers, the net intensity $\left({I}_{n}\right)$ of the emergent light can be written as ${I}_{n}=\dfrac{I}{{2}^{n}}......................(1)$ Comparing equation (1) with the given expression for the intensity, it can be concluded that $n=6$
5. Q5JEE Main 2022 (29 Jul, Shift 1)Interference and Superposition of Waves
   Two light beams of intensities $4I$ and $9I$ interfere on a screen. The phase difference between these beams on the screen at point $A$ is zero and at point $B$ is $\pi$. The difference of resultant intensities, at the point $A$ and $B$, will be _____ $I$.
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   Answer: 24

   The resultant intensity is given by ${I}_{net}={I}_{1}+{I}_{2}+2\sqrt{{I}_{1}}\sqrt{{I}_{2}}\cos \phi$ Now, for ${I}_{\max }$, phase angle is $\phi =0$ and for ${I}_{\min }$, phase angle is $\phi =\pi$. Thus, ${I}_{\max }={\left(\sqrt{{I}_{1}}+\sqrt{{I}_{2}}\right)}^{2}={\left(\sqrt{9I}+\sqrt{4I}\right)}^{2}=25I$ And ${I}_{\min }={\left(\sqrt{{I}_{1}}-\sqrt{{I}_{2}}\right)}^{2}={\left(\sqrt{9I}-\sqrt{4I}\right)}^{2}=I$ The difference of resultant intensities is ${I}_{\max }-{I}_{\min }=25I-I=24I$