Question

In a YDSE arrangement separation between slits is d, screen is at distance D from the slits, and $\lambda$ is wavelength of light used.If intensity of light from each slit on the screen is $I_0$, then the minimum seperation between two lines on the screen, where intensity is $I_0$, is

• A. $\frac{2\lambda D}{3d}$
• B. $\frac{\lambda D}{3d}$
• C. $\frac{3\lambda D}{4d}$
• D. $\frac{\lambda D}{2d}$

Answer 1

The correct option is B $\frac{\lambda D}{3d}$

$I=4I_0{\cos }^2\left(\frac{\delta }{2}\right)\Rightarrow I_0=4I_0{\cos }^2\left(\frac{\delta }{2}\right)\Rightarrow \delta =\frac{2\pi }{3}\because \delta =\frac{2\pi }{3}\times \Delta x\Rightarrow \frac{2\pi }{3}=\frac{2\pi }{\lambda }\left(y\frac{d}{D}\right)\Rightarrow y=\frac{\lambda D}{3d}\therefore (\Delta y{)}_{min}=2\times (\frac{\lambda D}{2d}-\frac{\lambda D}{3d})=\frac{\lambda D}{3d}$