JEE Mains · Physics · STD 12 - 10. Wave optics
As shown in the figure, in Young's double slit experiment, a thin plate of thickness \(t =10\,\mu m\) and refractive index \(\mu _{1}=1.2\) is inserted infront of slit \(S _1\). The experiment is conducted in air \((\mu=1)\) and uses a monochromatic light of wavelength \(\lambda=500\,nm\). Due to the insertion of the plate, central maxima is shifted by a distance of \(x \beta_0 . \beta_0\) is the fringe-width before the insertion of the plate.The value of the \(x\) is \(.............\)
- A \(2\)
- B \(3\)
- C \(6\)
- D \(4\)
Answer & Solution
Correct Answer
(D) \(4\)
Step-by-step Solution
Detailed explanation
\(\text { Fringe shift }=\frac{ t (\mu-1)}{\lambda} B\) \(=\frac{10 \times 10^{-6}(1.2-1)}{5 \times 10^{-7}} B\) \(=\frac{10^{-5} \times 0.2}{5 \times 10^{-7}}=4\)
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