A parallel beam of monochromatic light of wavelength 600 nm passes through single slit of 0.4 mm width. Angular divergence corresponding to second order minima would be _________ × 10⁻³ rad. Numerical answer type. Work it out, then open the answer.

Answer: 6. Minima of a single slit: asin θ = nλ, so for small angles θn = (nλ)/(a). Second minimum: θ₂ = 2 × 600×10⁻⁹0.4×10⁻³ = 3×10⁻³ rad on each side. Angular divergence between the second minima on both sides = 2θ₂ = 6×10⁻³ rad. So the answer is 6.

A parallel beam of monochromatic light of wavelength 600 nm passes… | JEE Main 2024 (8 Apr, Shift 1)

Q25JEE Main 2024 · 8 APR · SHIFT 1Easy

Diffraction at a Single Slit

A parallel beam of monochromatic light of wavelength $600\text{ nm}$ passes through single slit of $0.4\text{ mm}$ width. Angular divergence corresponding to second order minima would be _________ $\times 10^{-3}\text{ rad}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 6

Minima of a single slit: $a\sin\theta = n\lambda$ , so for small angles $\theta_n = \dfrac{n\lambda}{a}$ .
Second minimum: $\theta_2 = \dfrac{2 \times 600\times10^{-9}}{0.4\times10^{-3}} = 3\times10^{-3}\text{ rad}$ on each side.
Angular divergence between the second minima on both sides $= 2\theta_2 = 6\times10^{-3}\text{ rad}$ .
So the answer is $6$ .

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