JEE Mains · Physics · STD 12 - 10. Wave optics
In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness \(t\) and refractive index \(n (=1.5)\), the central fringe shifts by 0.2 cm . The value of t is _________ cm .
- A \(8 \times 10^{-4}\)
- B \(6.0 \times 10^{-3}\)
- C \(5.6 \times 10^{-4}\)
- D \(5.0 \times 10^{-3}\)
Answer & Solution
Correct Answer
(A) \(8 \times 10^{-4}\)
Step-by-step Solution
Detailed explanation
\(d \sin \theta=(\mu-1) t\) \(d \left[\frac{ x }{ D }\right]=(\mu-1) t\) \(t=\frac{x d}{D(\mu-1)}\) \(=\frac{(0.2)(0.1)}{50(1.5-1)}\) \(t =8 \times 10^{-4} cm\)
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