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JEE ADVANCED 2025
Paper-2 2025
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 = 2mm and the shortest distance between the slits and the screen is D = 2m. Thickness of the combination of the wedges is t = 12m. The value of  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 0 by ____.
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Solution

$\begin{aligned} & \left(x_1\right)_{\text {extra }}=\left[(1.5-1)^3+(1.7-1) 9\right] \times 10^{-6} \\ & =(1.5+6.3) \times 10^{-6} \\ & =7.8 \times 10^{-6} \\ & \left(x_2\right)_{\text {extra }}=[(1.5-1) 9+(1.7-1) 3] \times 10^{-6} \\ & =(4.5+2.1) \times 10^{-6} \\ & =6.6 \times 10^{-6} \\ & \Delta x=1.2 \times 10^{-6} \\ & \frac{d}{D} y=1.2 \times 10^{-6} \\ & y=\frac{1.2 \times 10^{-6} \times 2}{2 \times 10^{-3}} \\ & =1.2 \mathrm{~mm}\end{aligned}$
Question Tags
JEE Advance
Physics
Easy
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