Question

In young's Double Slit Experiment, the distance between the slits and the screen is 1.2 m and the distance between the two slits is 2.4 mm. If a thin transparent mica sheet of thickness \(1 \mu m\) and R.I. 1.5 is introduced between one of the interfering beams, the shift in the position of central bright fringe is

• A. 2 mm
• B. 0.5 mm
• C. 0.125 mm
• D. 0.25 mm

Answer 1

The Correct Option is D

Path difference due to insertion of mica sheet $\Delta x =(\mu-1) t$
Let the shift in the fringe pattern be $'y'$
Also, path difference $\Delta x=y \times d / D$,
so comparing both $(\mu-1) t=y \times(d / D)$
$y =(\mu-1) t \times( D / d )$,
where $\mu=1.5, D =2.4$ and $d =1.2$
putting the values, we get
$y =0.25 \,mm$.

Answer 2

In Young's Double Slit Experiment, the introduction of a mica sheet causes a change in the optical path length, which results in a shift in the position of the central bright fringe. The shift is calculated using the formula: \[ \Delta y = \frac{t \cdot \mu - 1}{d} \cdot L \] Where:
\(t = 1 \, \mu m = 1 \times 10^{-6} \, m\) is the thickness of the mica sheet,
\(\mu = 1.5\) is the refractive index of the mica,
\(d = 2.4 \, mm = 2.4 \times 10^{-3} \, m\) is the distance between the slits,
\(L = 1.2 \, m\) is the distance between the slits and the screen.
Substituting the values: \[ \Delta y = \frac{(1 \times 10^{-6} \, m) \cdot (1.5 - 1)}{2.4 \times 10^{-3} \, m} \cdot 1.2 \, m \] \[ \Delta y = \frac{0.5 \times 10^{-6}}{2.4 \times 10^{-3}} \cdot 1.2 = 0.25 \, mm \]
Thus, the shift in the position of the central bright fringe is 0.25 mm, and the correct answer is (D).