The diffraction pattern of a light of wavelength 400 nm diffracting from a slit of width 0.2 mm is focused on the focal plane of a convex lens of focal length 100 cm. The width of the 1 st secondary maxima will be: (a) 2 mm (b) 0.2 mm (c) 0.02 mm (d) 2 cm

Answer: (a). A secondary maximum lies between two minima that are (λ f)/(a) apart, so its width = (λ f)/(a). = 400 × 10⁻⁹ × 10.2 × 10⁻³ = 2 × 10⁻³ m So the answer is (a) 2 mm.

The diffraction pattern of a light of wavelength 400 nm diffracting… | JEE Main 2024 (30 Jan, Shift 1)

Q39JEE Main 2024 · 30 JAN · SHIFT 1Medium

Diffraction at a Single Slit

The diffraction pattern of a light of wavelength $400\text{ nm}$ diffracting from a slit of width $0.2\text{ mm}$ is focused on the focal plane of a convex lens of focal length $100\text{ cm}$ . The width of the $1$ st secondary maxima will be:

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Answer: (a)

A secondary maximum lies between two minima that are $\dfrac{\lambda f}{a}$ apart, so its width $= \dfrac{\lambda f}{a}$ .
$= \dfrac{400 \times 10^{-9} \times 1}{0.2 \times 10^{-3}} = 2 \times 10^{-3}$ m
So the answer is (a) $2$ mm.

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