Question

In YDSE, if $D$ equals distance of screen and $d$ is separation between the slit. The distance of nearest point to the central maximum where intensity is same as that due to a single slit is equal to

• A. $\frac{D\lambda }{d}$
• B. $\frac{D\lambda }{2d}$
• C. $\frac{D\lambda }{3d}$
• D. $\frac{3D\lambda }{d}$

Answer 1

Answer: (C)

$\frac{D\lambda }{3d}$

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