JEE Advance - Physics (2025 - Paper 2 Online - No. 14)
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 ____________.
Answer
1.20
Explanation
$\begin{aligned} & x+y=12 \mu m \\ & \frac{4}{12}=\frac{1}{x} \\ & x=3 \mu m \\ & y=6 \mu m\end{aligned}$
$\begin{aligned} & \therefore \Delta=\frac{y d}{D}-\left(\mu_B-1\right) x-\left(\mu_A-1\right) y+\left(\mu_B-1\right) y+\left(\mu_A-1\right) x \\ & \frac{-y d}{D}=-0.5 \times 3-0.7 \times 9+0.5 \times 9+0.7 \times 3 \\ & \frac{-y d}{D}=-0.5 \times 6-0.7 \times 6 \\ & \Rightarrow \frac{-y d}{D}=-1.2 \mu \mathrm{~m} \\ & \Rightarrow y=\frac{1.2 \times D}{d}=\frac{1.2 \times 2}{2 \times 10^{-3}} \times 10^{-6} \\ & =1.2 \mathrm{~mm}\end{aligned}$
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