A single slit of width $a = 0.06 mm$ is illuminated by monochromatic light of wavelength $λ = 550 nm$ . Find the angle $θ$ (in degrees) of the first minimum in the diffraction pattern.

easy 1 mark Class 12 · Physics · Wave Optics

Answer

θ \approx 0.5 3^{\circ}

The correct answer is $θ \approx 0.5 3^{\circ}$ — Compute $θ = arcsin (9.17 \times 1 0^{- 3}) \approx 0.52 6^{\circ}$ , rounded to two decimal places.

Step-by-step solution

Apply the condition for minima in single‑slit diffraction: $; a s in θ = m l amb d a$ , where $m = 1$ for the first minimum.
Convert the slit width to metres: $a = 0.06 mm = 6 \times 1 0^{- 5} m$ .
Convert the wavelength to metres: $λ = 550 nm = 5.5 \times 1 0^{- 7} m$ .
Insert the values into $a s in θ = λ$ : $s in θ = \frac{5.5 \times 1 0 ^{- 7}}{6 \times 1 0 ^{- 5}} = 9.17 \times 1 0^{- 3}$ .
Compute $θ = arcsin (9.17 \times 1 0^{- 3}) \approx 0.52 6^{\circ}$ , rounded to two decimal places.

Common mistake: Using

a s in θ = m l amb d a /2

instead of

m l amb d a

Concept tested: Single‑slit diffraction minima

Published 4 October 2026 · How this solution was prepared and checked

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A single slit of width a=0.06 mm is illuminated by monochromatic light of wavelength λ=550 nm. Find the angle θ (in degrees) of the first minimum in the diffraction pattern.

Step-by-step solution

A single slit of width $a = 0.06 mm$ is illuminated by monochromatic light of wavelength $λ = 550 nm$ . Find the angle $θ$ (in degrees) of the first minimum in the diffraction pattern.