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 minima is at ${60}^o$ from the central maxima. If the first minimum is produced at ${\theta }_1$, then ${\theta }_1$ is close to

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

Answer 1

The correct option is D ${25}^o$

For single slit diffraction experiment:
Angle of minima are given by
$$\begin{array}{cc}& \sin {\theta }_n=\frac{n\lambda }{d}(\sin {\theta }_n\ne {\theta }_n\text{as}\theta \text{is large})\\ & \sin {\theta }_2=\sin {60}^{\circ }=\frac{\sqrt{3}}{2}=\frac{2\lambda }{d}=\frac{2\times 6000\times {10}^{-10}}{d}& & \sin {\theta }_1=\frac{\lambda }{d}=\frac{6000\times {10}^{-10}}{d}& & \text{Dividing (1) and (2)}\\ & ⟹\frac{\sqrt{3}}{2sin{\theta }_1}=2⟹sin{\theta }_1=\frac{\sqrt{3}}{4}=0.43\\ \text{(1)}& \text{(2)}\end{array}$$
As, the value is coming less than ${30}^o$ the only available option are ${20}^o$ and ${25}^o$ but by using approximation we get ${\theta }_1={25}^{\circ }$