Question

The angular deviation of $5th$ 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 $5th$ 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 $\lambda =3768\mathring{A}$