Question

In YDSE experiment distance on screen from central bright fringe where intensity is $25\%$ of maximum intensity is

• A. $\frac{D\lambda }{3d}$
• B. $\frac{4\lambda D}{3d}$
• C. $\frac{D\lambda }{\sqrt{2}d}$
• D. $\frac{7\lambda D}{3d}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $\frac{D\lambda }{3d}$
B $\frac{4\lambda D}{3d}$
D $\frac{7\lambda D}{3d}$


$I=I_{0}co{s}^2\frac{\varphi }{2}\Rightarrow \frac{I_0}{4}=I_{0}co{s}^2\frac{\varphi }{2}$

$\frac{\varphi }{2}=\frac{\pi }{3}\Rightarrow$ By $\varphi =\frac{2\pi }{\lambda }\Delta x\Rightarrow \Delta x=\frac{\lambda }{3}$

$\frac{\lambda }{3}=d\frac{y}{D}\Rightarrow y=\frac{D\lambda }{3d}=y_0$

Fringe width $w=\frac{\lambda D}{d}$

So $I=\frac{I_0}{4}$ at $y=y_0+nw$
For $n=0,1\text{and}2$,
$y=\frac{D\lambda }{3d}$,
$y=\frac{4D\lambda }{3d}$ and
$y=\frac{7\lambda D}{3d}$ respectively