Question

In a YDSE, light of wavelength \(\lambda = 5000\,\text{\AA}\) is used, which emerges in phase from two slits at a distance \(d = 3\times10^{-7}\,\text{m}\) apart. A transparent sheet of thickness \(t = 1.5\times10^{-7}\,\text{m}\) and refractive index \(\mu = 1.17\) is placed over one of the slits. What is the angular position of the central maxima of the interference pattern from the centre of the screen? Find the value of \(y\). 

• A. \(4.9^\circ\) and \(\dfrac{D(\mu-1)t}{2d}\)
• B. \(4.9^\circ\) and \(\dfrac{D(\mu-1)t}{d}\)
• C. \(3.9^\circ\) and \(\dfrac{D(\mu+1)t}{d}\)
• D. \(2.9^\circ\) and \(\dfrac{2D(\mu+1)t}{d}\)

Hint:

Insertion of a thin sheet shifts the central maxima: \[ \Delta y = D\frac{(\mu-1)t}{2d} \] Angular position depends only on path difference.

Answer 1

The Correct Option is A

Step 1: Extra path introduced by sheet: \[ \Delta = (\mu-1)t \]
Step 2: Condition for central maxima: \[ d\sin\theta = \frac{(\mu-1)t}{2} \]
Step 3: Substituting values: \[ \sin\theta = \frac{(1.17-1)\times1.5\times10^{-7}}{2\times3\times10^{-7}} \approx 0.085 \]
Step 4: Hence: \[ \theta \approx 4.9^\circ \]
Step 5: Linear shift on screen: \[ y = D\frac{(\mu-1)t}{2d} \]