Question 47: Waves Optics

Question

In a Young's double slit experiment, a combination of two glass wedges $A$ and $B$, having refractive indices 1.7 and 1.5, respectively, are placed in front of the slits, as shown in the figure. The separation between the slits is $d=2 \mathrm{~mm}$ and the shortest distance between the slits and the screen is $D=2 \mathrm{~m}$. Thickness of the combination of the wedges is $t=12 \mu \mathrm{~m}$. The value of $l$ as shown in the figure is 1 mm . Neglect any refraction effect at the slanted interface of the wedges. Due to the combination of the wedges, the central maximum shifts (in mm ) with respect to O by ____________.

Accepted Answer

$\begin{aligned} & x+y=12 \mu m \ & \frac{4}{12}=\frac{1}{x} \ & x=3 \mu m \ & y=6 \mu m\end{aligned}$
$\begin{aligned} & 	herefore \Delta=\frac{y d}{D}-\left(\mu_B-1ight) x-\left(\mu_A-1ight) y+\left(\mu_B-1ight) y+\left(\mu_A-1ight) x \ & \frac{-y d}{D}=-0.5 	imes 3-0.7 	imes 9+0.5 	imes 9+0.7 	imes 3 \ & \frac{-y d}{D}=-0.5 	imes 6-0.7 	imes 6 \ & \Rightarrow \frac{-y d}{D}=-1.2 \mu \mathrm{~m} \ & \Rightarrow y=\frac{1.2 	imes D}{d}=\frac{1.2 	imes 2}{2 	imes 10^{-3}} 	imes 10^{-6} \ & =1.2 \mathrm{~mm}\end{aligned}$

Waves Optics

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