Wave Optics

JEE Main level, single correct, medium

144 of 191

Question 144 of 191

Light of wavelength 550 nm550 \mathrm{~nm} falls normally on a slit of width 22.0×10−5 cm22.0 \times 10^{-5} \mathrm{~cm}. The angular position of the second minima from the central maximum will be (in radians)

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Answer: (A) π8\frac{\pi}{8}

If angular position of 2nd 2^{\text {nd }} maxima from central maxima is θ\theta then
sin⁡θ=(2n−1)λ2a=3λ2a=3×550×10−92×22×10−7∴θ=π8rad\begin{aligned} &\sin \theta=\frac{(2 \mathrm{n}-1) \lambda}{2 \mathrm{a}}=\frac{3 \lambda}{2 \mathrm{a}}=\frac{3 \times 550 \times 10^{-9}}{2 \times 22 \times 10^{-7}} \\ &\therefore \quad \theta=\frac{\pi}{8} \mathrm{rad} \end{aligned}

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