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JEE Mains · Physics · STD 12 - Wave optics
  • JEE Mains 2026
  • Easy
  • MCQ

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 .

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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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