Question

In standard Young's double slit experiment, visible light $(390nm,700nm)$ is used in both the slits. Distance between the slits is $2mm$, and distance of screen from the slits is $1m$. A thin glass slab ($\mu =$ 1.5) of thickness $6$ $\mu m$ is placed in front of slit $S_2$. Light of which wavelength (s) is missing at point $P$, which is directly in front of slit $S_2$. (Neglect dispersion of light)

• A. $400nm$
• B. $500nm$
• C. $\frac{2000}{3}nm$
• D. $\frac{1600}{3}nm$

Answer 1

Answer: (A), (C)

The correct options are
A $400nm$
C $\frac{2000}{3}nm$

We have the net path difference,
$=(\mu -1)t-\frac{yd}{D}=(\frac{3}{2}-1)6\mu m-\frac{\left(d/2\right)d}{D}=1\mu m$

For destructive interference we have,
$\Delta x=1\mu m=(2n+1)\frac{\lambda }{2}$
$⟹\lambda =\frac{2}{(2n+1)}\mu m$
Hence, $\lambda =2000nm,\frac{2000}{3}nm,\frac{2000}{5}nm$