Question

Visible light of wavelength $6000 \times 10^{-8}$ cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at $60^{\circ}$ from the central maximum. If the first minimum is produced at $\theta_{1}$, then $\theta_{1}$ is close to :

• A. 45$^{\circ}$
• B. 30$^{\circ}$
• C. 25$^{\circ}$
• D. 20$^{\circ}$

Answer 1

Answer: (C)

For $2^{nd}$ minima
$d\,sin\theta=2\lambda$
$sin\theta=\frac{\sqrt{3}}{2}$ (given)
$\Rightarrow \frac{\lambda}{d}=\frac{\sqrt{3}}{4}\,...\left(i\right)$
So for $1^{st}$ minima is
$d\,sin\theta=\lambda$
$sin\theta =\frac{\lambda}{d}=\frac{\sqrt{3}}{4}$ (from equation (i))
$\theta=25.65^{\circ}$ (from sin table)
$\theta\approx25^{\circ}$