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
$dsin\theta =2\lambda$
$sin\theta =\frac{\sqrt{3}}{2}$ (given)
$\Rightarrow \frac{\lambda }{d}=\frac{\sqrt{3}}{4}...\left(i\right)$
So for $1^{st}$ minima is
$dsin\theta =\lambda$
$sin\theta =\frac{\lambda }{d}=\frac{\sqrt{3}}{4}$ (from equation (i))
$\theta =25.65^{\circ }$ (from sin table)
$\theta \approx 25^{\circ }$