Question

In the Young’s double slit experiment apparatus shown in figure, the ratio of maximum to minimum intensity on the screen is 9. The wavelength of light used is $\lambda$, then the value of y is

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

Answer 1

The correct option is C

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

$\frac{I_{max}}{I_{min}}={\left(\frac{\sqrt{\frac{I_1}{I_2}+1}}{\sqrt{\frac{I_1}{I_2}-1}}\right)}^2=\frac{9}{1}$

or $\frac{x+1}{x-1}=3(x=\sqrt{\frac{I_1}{I_2}})$

$\therefore x=2$

$\therefore \frac{I_1}{I_2}=4$

$I_1=4I_2$

i.e., if $I_2=I_0$

then $I_1=4I_0$

$I_0=4I_0=co{s}^2\frac{\varphi }{2}$

$\therefore \varphi =\frac{2\pi }{3}$

$\therefore \left(\frac{2\pi }{\lambda }\right)\left(\frac{yd}{D}\right)=\frac{2\pi }{3}$

$\therefore y=\frac{\lambda D}{3d}$