JEE Advanced · Physics · 25. Wave Optics
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 ______
- A 1.1
- B 1.2
- C 1.3
- D 1.4
Answer & Solution
Correct Answer
(B) 1.2
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & x+y=12 \mu m \\ & \frac{4}{12}=\frac{1}{x} \\ & x=3 \mu m \\ & y=6 \mu m\end{aligned}\)
\(\therefore \Delta=\frac{y d}{D}-\left(\mu_B-1\right) x-\left(\mu_A-1\right) y~+\) \(\left(\mu_B-1\right) y+\left(\mu_A-1\right) x \)
\( \frac{-y d}{D}=-0.5 \times 3-0.7 \times 9+0.5 \times 9+0.7 \times 3 \)
\( \frac{-y d}{D}=-0.5 \times 6-0.7 \times 6 \)
\( \Rightarrow \frac{-y d}{D}=-1.2 \mu \mathrm{~m} \)
\( \Rightarrow y=\frac{1.2 \times D}{d}=\frac{1.2 \times 2}{2 \times 10^{-3}} \times 10^{-6} \)
\( =1.2 \mathrm{~mm}\)
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