Q 12-10-074JEE MainJEE Main 2023 (29 Jan, Shift 1)Medium
In a Young's double slit experiment, two slits are illuminated with a light of wavelength $800\ \text{nm}$. The line joining $A_1P$ is perpendicular to $A_1A_2$ as shown in the figure. If the first minimum is detected at $P$, the value of slits separation $a$ will be
The distance of screen from slits $D=5\ \text{cm}$.
Answer: (C) $0.2\ \text{mm}$
$A_1P=D$ and $A_2P=\sqrt{D^2+a^2}$. For the first minimum the path difference is $\lambda/2$:
$$\sqrt{D^2+a^2}-D=\frac{\lambda}{2}\ \Rightarrow\ a^2=D\lambda+\frac{\lambda^2}{4}\approx D\lambda$$
$$a=\sqrt{0.05\times800\times10^{-9}}=\sqrt{4\times10^{-8}}=2\times10^{-4}\ \text{m}=0.2\ \text{mm}$$
Solution by Sreeraj P, M.Sc Physics