JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2026
- Hard
- Numerical
In a Young's double slit experiment set up, the two slits are kept 0.4 mm apart and screen is placed at 1 m from slits. If a thin transparent sheet of thickness \(20 \mu m\) is introduced in front of one of the slits then centre bring fringe shifts by 20 mm on the screen. The refractive index of transparent sheet is given by \(\frac{\alpha}{10}\), where \(\alpha\) is _________ .
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) 14
Step-by-step Solution
Detailed explanation
\(y _{\text {shift }}=\frac{(\mu-1) t \cdot D }{ d }\)
\(20910^{-3}=\frac{(\mu-1) \times 20 \times 10^{-6} \times 1}{0.4 \times 10^{-3}}\)
\((\mu-1) \Rightarrow 0.4\)
\(\mu \Rightarrow 1.4\)
\(\frac{\alpha}{10}=1.4, \alpha=14\)
\(20910^{-3}=\frac{(\mu-1) \times 20 \times 10^{-6} \times 1}{0.4 \times 10^{-3}}\)
\((\mu-1) \Rightarrow 0.4\)
\(\mu \Rightarrow 1.4\)
\(\frac{\alpha}{10}=1.4, \alpha=14\)
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