Light of wavelength 550 nm falls normally on a slit of width 22.0 × 10⁻⁵ cm. The angular position of the second minima from the central maximum will be (in radians): (a) π/12 (b) π/8 (c) π/6 (d) π/4

Answer: (c). Second minimum: asin θ = 2λ. sin θ = 2× 550×10⁻⁹22×10⁻⁷ = 0.5 ⇒ θ = π/6. So the answer is (c) π/6.

Light of wavelength 550 nm falls normally on a slit of width 22.0 ×… | JEE Main 2018 (15 Apr, Shift 1)

Q72JEE Main 2018 · 15 APR · SHIFT 1Easy

Diffraction at a Single Slit

Light of wavelength $550\text{ nm}$ falls normally on a slit of width $22.0 \times 10^{-5}\text{ cm.}$ The angular position of the second minima from the central maximum will be (in radians):

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Answer: (c)

Second minimum: $a\sin\theta = 2\lambda$ .
$\sin\theta = \frac{2\times 550\times10^{-9}}{22\times10^{-7}} = 0.5 \Rightarrow \theta = \frac{\pi}{6}$ .
So the answer is (c) $\frac{\pi}{6}$ .

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Diffraction at a Single Slit question from JEE Main 2018 (15 Apr, Shift 1)

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