Question

In a Young's double slit experiment set up, the two slits are kept 0.4 mm apart and screen is placed at 1 m from slits. If a thin transparent sheet of thickness 20 $\mu$m is introduced in front of one of the slits then center bright fringe shifts by 20 mm on the screen. The refractive index of transparent sheet is given by $\frac{\alpha{10}$, where $\alpha$ is _________}.

Hint:

The fringe shift formula can be easily remembered by noting that $(\mu-1)t$ is the path difference. Multiplying by $D/d$ converts path difference to linear displacement on the screen.

Answer 1

Correct Answer: 15

Step 1: Understanding the Concept:
When a transparent sheet is placed in front of one slit in a Young's Double Slit Experiment (YDSE), it introduces an additional optical path length.
This causes the entire fringe pattern to shift. The central bright fringe (zeroth-order maximum) shifts to a position where the path difference created by the geometry of the slits compensates for the path difference introduced by the sheet.

Step 2: Key Formula or Approach:
The shift in the fringe pattern ($\Delta y$) is given by the formula:
\[ \Delta y = \frac{D}{d} (\mu - 1) t \]
where:
$D$ = Distance to the screen
$d$ = Separation between the slits
$\mu$ = Refractive index of the transparent sheet
$t$ = Thickness of the sheet

Step 3: Detailed Explanation:
From the question, we have the following parameters:
$d = 0.4$ mm $= 0.4 \times 10^{-3}$ m
$D = 1$ m
$t = 20$ $\mu$m $= 20 \times 10^{-6}$ m $= 2 \times 10^{-5}$ m
$\Delta y = 20$ mm $= 20 \times 10^{-3}$ m $= 2 \times 10^{-2}$ m
Rearranging the shift formula to solve for $(\mu - 1)$:
\[ \mu - 1 = \frac{\Delta y \cdot d}{D \cdot t} \]
Substitute the known values:
\[ \mu - 1 = \frac{(2 \times 10^{-2}) \times (4 \times 10^{-4})}{1 \times (2 \times 10^{-5})} \]
\[ \mu - 1 = \frac{8 \times 10^{-6}}{2 \times 10^{-5}} = 4 \times 10^{-1} = 0.4 \]
Thus, the refractive index is:
$\mu = 1 + 0.4 = 1.4$
The problem states that the refractive index is given by $\frac{\alpha}{10}$:
$\frac{\alpha}{10} = 1.4 \Rightarrow \alpha = 14$

Step 4: Final Answer:
The value of $\alpha$ is 14.