Search any question & find its solution
Question:
Answered & Verified by Expert
In Young's double slit experiment, light of wavelength $\lambda$ passes through the double-slit and forms interference fringes on a screen $1.2 \mathrm{~m}$ away. If the difference between 3 rd order maximum and 3 rd order minimum is $0.18 \mathrm{~cm}$ and the slits are $0.02 \mathrm{~cm}$ apart, then $\lambda$ is
Options:
Solution:
2646 Upvotes
Verified Answer
The correct answer is:
$600 \mathrm{~nm}$
Distance between 3 rd order minima and 3 rd order maxima is $\frac{\beta}{2}$
$
\begin{array}{l}
\frac{\beta}{2}=0.18 \mathrm{~cm} \\
\beta=0.36 \mathrm{~cm} \\
\beta=\frac{\mathrm{D} \lambda}{\mathrm{d}} \\
\therefore \lambda=600 \mathrm{~nm}
\end{array}
$
$
\begin{array}{l}
\frac{\beta}{2}=0.18 \mathrm{~cm} \\
\beta=0.36 \mathrm{~cm} \\
\beta=\frac{\mathrm{D} \lambda}{\mathrm{d}} \\
\therefore \lambda=600 \mathrm{~nm}
\end{array}
$
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.