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JEE Main 2-April-2025 Paper - Shift 1 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 2-April-2025 Paper - Shift 14 Marks
Question
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 $30^{\circ}$ in a single slit diffraction pattern recorded using 628 nm light, then the width of the slit is __________ $\mu m$.
✓

Answer

(6)
Explanation:
Image
$\theta_1=\sin ^{-1}\left(\frac{2 \lambda}{a}\right)$
$\theta_2=\sin ^{-1}\left(\frac{3 \lambda}{a}\right)$
$\because \quad \theta_1+\theta_2=30^{\circ}$
$\Rightarrow \sin ^{-1}\left(\frac{2 \lambda}{ a }\right)+\sin ^{-1}\left(\frac{3 \lambda}{ a }\right)=\frac{\pi}{6}$
$\Rightarrow \frac{2 \lambda}{ a } \sqrt{1-\left(\frac{3 \lambda}{ a }\right)^2}+\frac{3 \lambda}{ a } \sqrt{1+\left(\frac{2 \lambda}{ a }\right)^2}=\sin \frac{\pi}{6}$
Here $\lambda=628 nm$
After solving$A=6.07 \mu m$
Approximate Method :
$\theta=\theta_1+\theta_2$
$\Rightarrow \frac{\pi}{6}=\frac{2 \lambda}{ a }+\frac{3 \lambda}{ a }$
$\Rightarrow \frac{\pi}{6}=\frac{5}{ a }(628 nm)$
$\Rightarrow a =6 \mu m$

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