Question

The central fringe in a Young’s double slit experiment with the wavelength $\lambda =500$ nm has intensity $I_0$. If one of the slits is covered by a $5\mu m$ thick film of glass $(\mu =1.5)$, the intensity at centre of the screen becomes $I_1$. The ratio of $\frac{I_1}{I_0}$ is

• A. $\frac{3}{16}$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

The correct option is D

$1$

Due to slit, extra path at centre = $t(\mu –1)$

$=5\times {10}^{-6}(1.5-1)$

$=\frac{5}{2}\times {10}^{-6}m=5\lambda$

Hence, $I_1=I_0$