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KCET · Physics · Wave Optics

In Young's double slit experiment, the distance between the slits and the screen is \(1.2 \mathrm{~m}\) and the distance between the two slits is \(2.4 \mathrm{~mm}\). If a thin transparent mica sheet of thickness \(1 \mu \mathrm{m}\) and RI \(1.5\) is introduced between one of the interfering beams, the shift in the position of central bright fringe is

  1. A \(2 \mathrm{~mm}\)
  2. B \(0.5 \mathrm{~mm}\)
  3. C \(0.125 \mathrm{~mm}\)
  4. D \(0.25 \mathrm{~mm}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(0.25 \mathrm{~mm}\)

Step-by-step Solution

Detailed explanation

Given, \(D=1.2 \mathrm{~m}, d=2.4 \mathrm{~mm}=2.4 \times 10^{-3} \mathrm{~m}\) \(t=1 \mu \mathrm{m}=1 \times 10^{-6} \mathrm{~m}\) and \(\mu=1.5\)
When a transparent sheet is introduced in the path of one of the interfering beam, the shift in the position of central bright fringe is
\(\begin{aligned}
y &=(\mu-1) t \frac{D}{d} \\
&=(1.5-1) 1 \times 10^{-6} \times \frac{1.2}{2.4 \times 10^{-3}} \\
&=\frac{0.5 \times 10^{-6} \times 1.2}{2.4 \times 10^{-3}} \\
&=0.25 \times 10^{-3} \text { or } 0.25 \mathrm{~mm}
\end{aligned}\)