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:

Start by writing the two path difference expressions directly: the shift caused by inserting the slab equals $(\mu-1)t$, and the shift measured on the screen equals $\dfrac{yd}{D}$, where d is the slit separation and D is the screen distance. Set these two expressions equal to each other and solve for t first, before touching the rest of the question. A common slip here is mixing centimetre and micrometre units for t and the shift, so convert everything to one consistent unit before you substitute the numbers.

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.

Answer 2

Concept:

• Inserting a slab in front of one slit adds an extra optical path of $(\mu-1)t$ on that side.
• The whole fringe pattern shifts to the point where this extra optical path is exactly cancelled by the usual geometric path difference $dy/D$, so we can build the shift formula from scratch instead of recalling it.

Step 1: Write the normal path difference on the screen.
For a point at distance $y$ from the centre, the geometric path difference between the two slits is $\Delta = \dfrac{dy}{D}$, where $d$ is the slit separation and $D$ is the slit-to-screen distance.

Step 2: Add the extra path from the slab.
The slab adds $(\mu-1)t$ to the path from the covered slit. The centre of the pattern moves to the new point $y'$ where the two effects cancel:
$\dfrac{dy'}{D} = (\mu - 1)t$

Step 3: Convert everything to SI units before substituting.
$y' = 0.2\ \text{cm} = 2\times10^{-3}\ \text{m}$, $d = 1\ \text{mm} = 1\times10^{-3}\ \text{m}$, $D = 50\ \text{cm} = 0.5\ \text{m}$, $\mu = 1.5$.

Step 4: Solve for $t$.
$t = \dfrac{y' \cdot d}{D(\mu-1)} = \dfrac{(2\times10^{-3})(1\times10^{-3})}{(0.5)(0.5)} = \dfrac{2\times10^{-6}}{0.25} = 8\times10^{-6}\ \text{m}$

Final Answer: $t = 8\ \mu m$