In Young’s Double Slit Experiment.... | JEE Main 2024

JEE MAIN 2024

05-04-24 S1

Question

In Young's double slit experiment, carried out with light of wavelength 5000 Å , the distance between the slits is 0.3 mm and the screen is at 200 cm from the slits. The central maximum is at x=0 cm. The value of x for third maxima is.............mm

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Solution

$$ \beta=\frac{\lambda D}{d}=\frac{5 \times 10^{-7} \times 2}{3 \times 10^{-4}}=\frac{10 \times 10^{-3}}{3} \mathrm{~m} $$ For $3^{\text {rd }}$ maxima $y_3=3 \beta=10 \times 10^{-3} \mathrm{~m}=10 \mathrm{~mm}$

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JEE Main

Physics

Medium

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