Question

In a double-slit experiment, \(6^{th}\) dark fringe is observed at a certain point of the screen. A transparent sheet of thickness \(t\) and refractive index \(n\) is now introduced in the path of one of the two interfering waves to increase its phase by \(2\pi(n-1)\,t/\lambda\). The pattern is shifted and \(8^{th}\) bright fringe is observed at the same point. Find the relation for thickness \(t\) in terms of \(n\) and \(\lambda\).

Hint:

Phase for 6th dark is \(11\pi\); for 8th bright it is \(16\pi\). The sheet adds \(5\pi\).

Answer 1

Step 1: Condition for the 6th dark fringe (without the sheet):
At a point on the screen, the phase difference between the two waves for the m-th dark fringe is \((2m - 1)\pi\). For the 6th dark fringe:
\[ \phi_1 = (2 \times 6 - 1)\pi = 11\pi \]

Step 2: Condition for the 8th bright fringe (with the sheet):
For the m-th bright fringe the phase difference is \(2m\pi\) (the central fringe is \(m = 0\)). For the 8th bright fringe:
\[ \phi_2 = 2 \times 8 \times \pi = 16\pi \]

Step 3: Use the extra phase from the sheet:
The sheet increases the phase of one wave by \(\dfrac{2\pi(n-1)t}{\lambda}\). This extra phase is the difference between the two situations:
\[ \frac{2\pi(n-1)t}{\lambda} = \phi_2 - \phi_1 = 16\pi - 11\pi = 5\pi \]

Step 4: Solve for t:
\[ (n-1)t = \frac{5\lambda}{2} \]
\[ t = \frac{5\lambda}{2(n-1)} \]

Final Answer:
The thickness of the sheet is \(t = \dfrac{5\lambda}{2(n-1)}\). \[ \boxed{t = \dfrac{5\lambda}{2(n-1)}} \]