The angular deviation of 5 th order dark fringe is $12^{\circ}$ in a single slit experiment. If the width of…

The angular deviation of 5 th order dark fringe is $12^{\circ}$ in a single slit experiment. If the width of the slit is $9 \mu \mathrm{m}$ then the wavelength of the incident light is
  1. $4862 Å$
  2. $5892 Å$
  3. $6002 Å$
  4. $3768 Å$

Solution

Given, angular deviation of 5th order dark
fringe, $\theta=12^{\circ}$,
Width of the slit, $d=9 \mu \mathrm{m}$
Now, angular deviation of $n$th fringe $=n \cdot \frac{\lambda}{d}$
$\begin{aligned}
\therefore 12 \times \frac{\pi}{180} & =5 \times \frac{\lambda}{9 \times 10^{-6}} \\
& \text { or } \quad \lambda=\frac{12 \times \pi}{180} \times \frac{9 \times 10^{-6}}{5} \mathrm{~m} \\
& \text { or } \quad \lambda=3768 Å
\end{aligned}$
So, the wave length of incident light is $\lambda=3768 Å$.

Asked in: MHT CET Full Test 9

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