Wave Optics

JEE Main level, numerical answer, medium

55 of 191

Question 55 of 191

In Young's double slit experiment, carried out with light of wavelength 5000A˚5000 \AA, the distance between the slits is 0.3 mm0.3 \mathrm{~mm} and the screen is at 200 cm200 \mathrm{~cm} from the slits. The central maximum is at x=0 cmx=0 \mathrm{~cm}. The value of xx for third maxima is _____mm\mathrm{mm}.

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Answer: 10

β=λDd=5×10−7×23×10−4=10×10−33 m\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 3rd 3^{\text {rd }} maxima y3=3β=10×10−3 m=10 mmy_3=3 \beta=10 \times 10^{-3} \mathrm{~m}=10 \mathrm{~mm}

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