Question

On replacing a thin film of mica of thickness $12 \times 10^{-5}$ cm in the path of one of the interfering beams in Young's double slit experiment using monochromatic light, the fringe pattern shifts through a distance equal to the width of bright fringe. If $\lambda = 6 \times 10^{-5}$ cm, the refractive index of mica is

• A. 1.1
• B. 1.3
• C. 1.5
• D. 1.4

Hint:

When the shift is equal to $n$ fringe widths, the optical path difference introduced is $(\mu - 1)t = n\lambda$.

Answer 1

The Correct Option is C

Step 1: Use the formula for the shift in the fringe pattern $\Delta x$ caused by a thin film of thickness $t$ and refractive index $\mu$. \[ \Delta x = \frac{D}{d}(\mu - 1)t \]
Step 2: Recall the expression for fringe width $\beta$ in Young's double slit experiment. \[ \beta = \frac{D\lambda}{d} \]
Step 3: The problem states the shift is equal to the width of one bright fringe, so $\Delta x = \beta$. \[ \frac{D}{d}(\mu - 1)t = \frac{D\lambda}{d} \]
Step 4: Simplify the equation by cancelling the common terms on both sides. \[ (\mu - 1)t = \lambda \]
Step 5: Solve for $\mu$. \[ \mu - 1 = \frac{\lambda}{t} \] \[ \mu = \frac{\lambda}{t} + 1 \]
Step 6: Substitute the values $\lambda = 6 \times 10^{-5}$ cm and $t = 12 \times 10^{-5}$ cm. \[ \mu = \frac{6 \times 10^{-5{12 \times 10^{-5 + 1 = 0.5 + 1 = 1.5 \]