Physics

Physics

Problem

A single slit of width bb is illuminated by a coherent monochromatic light of wavelength λ\lambda . If the second and fourth minima in the diffraction pattern at a distance 1 m from the slit are at 3 cm and 6 cm respectively from the central maximum, what is the width of the central maximum? (i.e. distance between first minimum on either side of the central maximum) a. 1.5 cm1.5 \text{ cm} b. 3.0 cm3.0 \text{ cm} c. 4.5 cm4.5 \text{ cm} d. 6.0 cm6.0 \text{ cm}
Intuitive explanation

Single-slit diffraction minima follow a simple rule: the mm-th minimum occurs at angle θm\theta_m where bsin⁡θm=mλb\sin\theta_m = m\lambda. The key insight is that you can use the two given minima to find both the slit width bb and wavelength λ\lambda simultaneously—then calculate where the first minimum lands. For small angles (valid here), sin⁡θ≈tan⁡θ≈y/D\sin\theta \approx \tan\theta \approx y/D, where yy is the position on screen and DD is the distance to it.

Intuitive explanation

Single-slit diffraction minima follow a simple rule: the mm-th minimum occurs at angle θm\theta_m where bsin⁡θm=mλb\sin\theta_m = m\lambda. The key insight is that you can use the two given minima to find both the slit width bb and wavelength λ\lambda simultaneously—then calculate where the first minimum lands. For small angles (valid here), sin⁡θ≈tan⁡θ≈y/D\sin\theta \approx \tan\theta \approx y/D, where yy is the position on screen and DD is the distance to it.

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