Search any question & find its solution
Question:
Answered & Verified by Expert
In a Young's double slit experiment, a light of wavelength $500 \mathrm{~nm}$ falls on it. Its slits separation is $2 \mathrm{~mm}$ and distance between plane of slits and screen is $2 \mathrm{~m}$ then, find distance of a point on the screen from central maxima where intensity becomes $50 \%$ of central maxima.
Options:
Solution:
2555 Upvotes
Verified Answer
The correct answer is:
$125 \mu \mathrm{m}$
$\begin{aligned} & \text { } I=I_0 \cos ^2 \frac{\phi}{2}=\frac{I_0}{2} \\ & \cos \frac{\phi}{2}=\frac{1}{\sqrt{2}} \Rightarrow \frac{\phi}{2}=\frac{\pi}{4}, \phi=\frac{\pi}{2} \\ & \Delta x=\frac{\lambda}{4} \\ & y=\frac{\beta}{4}=\frac{\lambda D}{4 d}=\frac{500 \times 10^{-9} \times 2}{4 \times 2 \times 10^{-3}}=125 \times 10^{-6} \mathrm{~m} \\ & y=125 \mu \mathrm{m}\end{aligned}$
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.