Question

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

Hint:

When solving YDSE problems with slabs, remember that the shift in the interference pattern depends on the refractive index and the thickness of the slab.

Answer 1

Correct Answer: 8

Step 1: Use the formula for path difference.
Path difference due to shift is neutralized by the path difference caused by the slab: \[ \frac{dy}{D} = (\mu - 1)t \] Where \( D \) is the distance between the slits and the screen, and \( \mu \) is the refractive index.
Step 2: Substitute the given values.
Given \( D = 50 \, \text{cm} \), \( y = 0.2 \, \text{cm} \), \( \mu = 1.5 \), and slit separation \( x = 1 \, \text{mm} \), we can solve for \( t \): \[ 10^{-3} \times 0.2 \times 10^{-2} = \frac{1}{2} t \] Simplifying, we get: \[ t = 8 \, \mu \text{m} \] Step 3: Conclusion.
The value of \( t \) is 8 \mu m.