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In Young's double slit experiment, carried out with light of wavelength $5000 \AA$, the distance between the slits is $0.3 \mathrm{~mm}$ and the screen is at $200 \mathrm{~cm}$ from the slits. The central maximum is at $x=0 \mathrm{~cm}$. The value of $x$ for third maxima is _____$\mathrm{mm}$.
PhysicsWave OpticsJEE MainJEE Main 2024 (05 Apr Shift 1)
Solution:
2723 Upvotes Verified Answer
The correct answer is: 10
$\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{5 \times 10^{-7} \times 2}{3 \times 10^{-4}}=\frac{10 \times 10^{-3}}{3} \mathrm{~m}$
For $3^{\text {rd }}$ maxima $y_3=3 \beta=10 \times 10^{-3} \mathrm{~m}=10 \mathrm{~mm}$

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