Question

The angular deviation of $5\, th$ order dark fringe is $12^{\circ}$ in a single slit experiment. If the width of the slit is $9 \,\mu m$ then the wavelength of the incident light is

• A. $4862 \,\mathring{A}$
• B. $5892 \,\mathring{A}$
• C. $6022 \,\mathring{A}$
• D. $3768 \,\mathring{A}$

Answer 1

Answer: (D)

Given, angular deviation of $5\, th$ order dark fringe, $\theta=12^{\circ}$,
Width of the slit, $d=9 \,\mu m$
Now, angular deviation of $n$ th fringe $=n \cdot \frac{\lambda}{d}$
$ \therefore 12 \times \frac{\pi}{180}=5 \times \frac{\lambda}{9 \times 10^{-6}}$
or $\lambda=\frac{12 \times \pi}{180} \times \frac{9 \times 10^{-6}}{5} m$
or $\lambda=3768\,\mathring{A}$
So, the wave length of incident light is $\lambda=3768 \,\mathring{A}$