2016 Wave OpticsSingle slit diffraction – minima and secondary maxima Medium Question 26 of 40 in Wave Optics

In a diffraction pattern due to a single slit of width a, the first minimum is observed at an angle 30° when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of

A $\sin^{-1}\left(\frac{2}{3}\right)$
B $\sin^{-1}\left(\frac{1}{2}\right)$
C $\sin^{-1}\left(\frac{3}{4}\right)$
D $\sin^{-1}\left(\frac{1}{4}\right)$

The answer is C.

Solution

First minimum at $\sin\theta = \lambda/a$; first secondary maximum at $\sin\theta' = 3\lambda/2a$.

For the first minimum: $a\sin\theta = \lambda \Rightarrow a = \frac{5\times10^{-7}}{\sin30^\circ} = 10\times10^{-7}$ m
For the first secondary maximum: $a\sin\theta' = \left(1 + \frac{1}{2}\right)\lambda$
$\sin\theta' = \frac{3}{2}\times\frac{5\times10^{-7}}{10\times10^{-7}} = \frac{3}{4}$
$\theta' = \sin^{-1}\left(\frac{3}{4}\right)$

Concept single-slit-secondary-maximum

Wave Optics in past papers

40 questions from this chapter have appeared across 18 exam years.