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JEE Main 20252 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Morning Shift
Subject
Physics
Chapter
Wave Optics

JEE Main 2025 · 2 April · Morning Shift · Physics · Wave Optics

If the measured angular separation between the second minimum to...

If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is 30∘30^{\circ} in a single slit diffraction pattern recorded using 628 nm light, then the width of the slit is _______ μm\mu \mathrm{m}.If the measured angular separation between the second minimum to the left of the central maximum and the third minimum

Correct answer

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Step-by-step solution

If the measured angular separation between the second minimum to the left of the central maximum and the third minimum
θ1=sin⁡−1(2λa)θ2=sin⁡−1(3λa)∵θ1+θ2=30∘\begin{aligned} & \theta_1=\sin ^{-1}\left(\frac{2 \lambda}{a}\right) \\ & \theta_2=\sin ^{-1}\left(\frac{3 \lambda}{a}\right) \\ & \because \quad \theta_1+\theta_2=30^{\circ}\end{aligned}
⇒sin⁡−1(2λa)+sin⁡−1(3λa)=π6⇒2λa1−(3λa)2+3λa1+(2λa)2=sin⁡π6\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 λ=628 nm\lambda=628 \mathrm{~nm}
After solving
A=6.07μ m\mathrm{A}=6.07 \mu \mathrm{~m}
Approximate Method :
θ=θ1+θ2⇒π6=2λa+3λa⇒π6=5a(628 nm)⇒a=6μ m\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}
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