Search any question & find its solution
Question:
Answered & Verified by Expert
In a Fraunhofer diffraction at a single slit of width $d$ and incident light of wavelength $5500 \mathrm{~A}$, the firs minimum is observed at an angle $30^{\circ}$. The first secondary maxima are observed at an angle $\theta=$
Options:
Solution:
2353 Upvotes
Verified Answer
The correct answer is:
$\sin ^{-1}\left(\frac{3}{4}\right)$
Slit width $=d$
$\lambda=5500 \AA=5.5 \times 10^{-7} \mathrm{~m}, \theta_1=30^{\circ}$
For first secondary minima, $d \sin \theta_1=\lambda$
$d=\frac{\lambda}{\sin \theta_1}=\frac{5.5 \times 10^{-7}}{\sin 30^{\circ}}=11 \times 10^{-7}$
For first secondary maxima, $d \sin \theta_1=\frac{3 \lambda}{2}$
$\begin{aligned} & \Rightarrow \sin \theta_1=\frac{3 \lambda}{2 d}=\frac{5.5 \times 10^{-7}}{2 \times 11 \times 10^{-7}}=\frac{3}{4} \\ & \theta_1=\sin ^{-1}\left(\frac{3}{4}\right)\end{aligned}$
$\lambda=5500 \AA=5.5 \times 10^{-7} \mathrm{~m}, \theta_1=30^{\circ}$
For first secondary minima, $d \sin \theta_1=\lambda$
$d=\frac{\lambda}{\sin \theta_1}=\frac{5.5 \times 10^{-7}}{\sin 30^{\circ}}=11 \times 10^{-7}$
For first secondary maxima, $d \sin \theta_1=\frac{3 \lambda}{2}$
$\begin{aligned} & \Rightarrow \sin \theta_1=\frac{3 \lambda}{2 d}=\frac{5.5 \times 10^{-7}}{2 \times 11 \times 10^{-7}}=\frac{3}{4} \\ & \theta_1=\sin ^{-1}\left(\frac{3}{4}\right)\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.