amanuj.s3.ap-south-1.amazonaws.com/pyq-question-images/batch-001/7e868fcb5a3e.webp" loading="lazy" style="max-width:100%;height:auto;display:block;margin:0 auto;max-height:40vh;vertical-align:baseline" alt="JEE Main 2025 (Online) 2nd April Morning Shift Physics - Wave Optics Question 33 English Explanation">

$$\begin{aligned} & \theta_1=\sin ^{-1}\left(\frac{2 \lambda}{a}\right) \\ & \theta_2=\sin ^{-1}\left(\frac{3 \lambda}{a}\right) \\ & \because \theta_1+\theta_2=30^{\circ} \end{aligned}$$

$$\begin{aligned} & \Rightarrow \sin ^{-1}\left(\frac{2 \lambda}{\mathrm{a}}\right)+\sin ^{-1}\left(\frac{3 \lambda}{\mathrm{a}}\right)=\frac{\pi}{6} \\ & \Rightarrow \frac{2 \lambda}{\mathrm{a}} \sqrt{1-\left(\frac{3 \lambda}{\mathrm{a}}\right)^2}+\frac{3 \lambda}{\mathrm{a}} \sqrt{1+\left(\frac{2 \lambda}{\mathrm{a}}\right)^2}=\sin \frac{\pi}{6} \end{aligned}$$

Here $\lambda=628 \mathrm{~nm}$

After solving

$$\mathrm{A}=6.07 \mu \mathrm{~m}$$

Approximate Method :

$$\begin{aligned} & \theta=\theta_1+\theta_2 \\ & \Rightarrow \frac{\pi}{6}=\frac{2 \lambda}{a}+\frac{3 \lambda}{a} \\ & \Rightarrow \frac{\pi}{6}=\frac{5}{a}(628 \mathrm{~nm}) \\ & \Rightarrow a=6 \mu \mathrm{~m} \end{aligned}$$

19:T5d0,

JEE Main 2025 (Online) 2nd April Morning Shift Physics - Wave Optics Question 33 Hindi Explanation

$$\begin{aligned} & \theta_1=\sin ^{-1}\left(\frac{2 \lambda}{a}\right) \\ & \theta_2=\sin ^{-1}\left(\frac{3 \lambda}{a}\right) \\ & \because \theta_1+\theta_2=30^{\circ} \end{aligned}$$

$$\begin{aligned} & \Rightarrow \sin ^{-1}\left(\frac{2 \lambda}{\math

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