In a YDSE set up, a slab of width t is inserted in front of one of slit. The interference pattern shifts by 0.2 cm on the screen. If the refractive index of slab is 1.5 than t is N µm (screen distance 50 cm and slits separation 1 mm) then N is _______

The correct answer is 8.

Understanding the Shift in YDSE:
When a transparent thin slab of thickness $t$ and refractive index $\mu$ is placed in front of one of the slits in a Young's Double Slit Experiment (YDSE), it introduces an additional optical path difference. This causes the entire interference pattern to shift on the screen.
The shift in the fringe pattern is given by the formula:
$\mathrm{\Delta y}=\frac{D}{d}(\mu -1)t$
where:
- $\mathrm{\Delta y}$ is the shift of the fringe pattern on the screen,
- $D$ is the distance between the slits and the screen,
- $d$ is the slit separation,
- $\mu$ is the refractive index of the slab,
- $t$ is the thickness of the slab.

Given Data:
- Shift, $\mathrm{\Delta y}=0.2\text{cm}=2\times {10}^{-3}\text{m}$
- Refractive index, $\mu =1.5$
- Screen distance, $D=50\text{cm}=0.5\text{m}$
- Slit separation, $d=1\text{mm}={10}^{-3}\text{m}$
- Thickness of the slab, $t=N\text{μm}=N\times {10}^{-6}\text{m}$

Step-by-Step Calculation:
Substitute the given values into the shift equation:
$2\times {10}^{-3}=\frac{0.5}{{10}^{-3}}\times (1.5-1)\times (N\times {10}^{-6})$
Simplify the terms step-by-step:
$2\times {10}^{-3}=500\times 0.5\times N\times {10}^{-6}$
$2\times {10}^{-3}=250\times N\times {10}^{-6}$
$2\times {10}^{-3}=2.5\times {10}^{-4}\times N$
Solving directly for $N$:
$N=\frac{2\times {10}^{-3}}{2.5\times {10}^{-4}}$
$N=\frac{20}{2.5}=8$