If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is 30° in a single slit diffraction pattern recorded using 628 nm light, then the width of the slit is _______ μ m. Numerical answer type. Work it out, then open the answer.

Answer: 6. n th minimum: θn ≈ (nλ)/(a) on either side of the centre. Second minimum on the left and third on the right: separation = (2λ)/(a) + (3λ)/(a) = (5λ)/(a) (5λ)/(a) = 30° = π/6 rad ⇒ a = (30λ)/(π) = 30 × 628 × 10⁻⁹3.14 a = 6 × 10⁻⁶ m = 6 μ m

If the measured angular separation between the second minimum to the… | JEE Main 2025 (2 Apr, Shift 1)

Q17JEE Main 2025 · 2 APR · SHIFT 1Medium

Diffraction at a Single Slit

If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is $30^\circ$ in a single slit diffraction pattern recorded using 628 nm light, then the width of the slit is _______ $\mu$ m.

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

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Answer: 6

$n$ th minimum: $\theta_n \approx \dfrac{n\lambda}{a}$ on either side of the centre.
Second minimum on the left and third on the right: separation $= \dfrac{2\lambda}{a} + \dfrac{3\lambda}{a} = \dfrac{5\lambda}{a}$
$\dfrac{5\lambda}{a} = 30^\circ = \dfrac{\pi}{6}$ rad $\Rightarrow a = \dfrac{30\lambda}{\pi} = \dfrac{30 \times 628 \times 10^{-9}}{3.14}$
$a = 6 \times 10^{-6}$ m $= 6\ \mu$ m

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