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 $60^{o}$ from the central maximum. If the first minimum is produced at $\theta _{1},$ then $\theta _{1}$ is close to

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

Answer 1

Answer: (C)

For $2^{n d}$ minima,
$dsin \theta =2\lambda $
$sin \theta =\frac{\sqrt{3}}{2}$ , (given)
$\Rightarrow \frac{\lambda }{d}=\frac{\sqrt{3}}{4}$ ...(1)
So, for $1^{s t}$ minima
$dsin\theta =\lambda $
$\Rightarrow sin\theta =\frac{\lambda }{d}=\frac{\sqrt{3}}{4}$
$\Rightarrow \theta =25.65^{o}$
$\Rightarrow \theta \approx25^{o}$