Question

In Young's double-slit experiment $d/D={10}^{-4}$ ($d=$ distance between slits, $D=$distance of screen from the slits). At point $P$ on the screen, resulting intensity is equal to the intensity due to the individual slit $I_0$. Then, the distance of point $P$ from the central maximum is ($\lambda =6000$ $\mathring{A}$):

• A. $2mm$
• B. $1mm$
• C. $0.5mm$
• D. $4mm$

Answer 1

The correct option is A $2mm$

At point P and C, intensities are $I_p=I_o,I_c={4I}_o$

$I_o={4I}_o{\cos }^2\frac{\Delta \varphi }{2}\Rightarrow \cos \frac{\Delta \varphi }{2}=\frac{1}{4}\Rightarrow \Delta \varphi =\frac{2\pi }{3}$

Distance from central maximum, $x=\Delta \varphi \frac{\beta }{2\pi }=\frac{\beta }{3}=\frac{\lambda D}{3d}$

So, $x=\frac{6\times {10}^{-7}}{3\times {10}^{-4}}=2\times {10}^{-3}m=2mm$