Q91 / 180 · AIIMS 2019 (26 May) · Physics
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.
- A$1000 \mu \mathrm{m}$
- B$500 \mu \mathrm{m}$
- C$250 \mu \mathrm{m}$
- Correct answerD$125 \mu \mathrm{m}$
Solution
$\begin{aligned} & \text { } I=I_0 \cos ^2 \frac{\phi}{2}=\frac{I_0}{2} \\ & \cos \frac{\phi}{2}=\frac{1}{\sqrt{2}} \Rightarrow \frac{\phi}{2}=\frac{\pi}{4}, \phi=\frac{\pi}{2} \\ & \Delta x=\frac{\lambda}{4} \\ & y=\frac{\beta}{4}=\frac{\lambda D}{4 d}=\frac{500 \times 10^{-9} \times 2}{4 \times 2 \times 10^{-3}}=125 \times 10^{-6} \mathrm{~m} \\ & y=125 \mu \mathrm{m}\end{aligned}$
Hence, the correct answer is option: 'D'.