Question

In a double slit experiment, when one of the slits is covered by a transparent mica sheet of refractive index 1.56, the central fringe shifts to the position of \( 7^{\text{th}} \) bright fringe, obtained with both slits uncovered. If the light source wavelength is 450 nm, the thickness of mica sheet is \( \alpha \times 10^{-9} \text{ m} \). The value of \( \alpha \) is _______.

Answer 1

Correct Answer: 5625

Step 1: Understanding the Concept:
Introducing a transparent sheet of thickness $t$ and refractive index $\mu$ in front of one slit introduces an additional optical path difference of $(\mu - 1)t$. This extra path difference shifts the entire interference pattern. If the shift corresponds to a specific fringe position, we can equate the path shift to the required wavelength condition.
Step 2: Key Formula or Approach:
Shift of the fringe pattern: $\Delta y = \frac{D}{d} (\mu - 1)t$
Position of the $n^{\text{th}}$ bright fringe (without sheet): $y_n = n \frac{\lambda D}{d}$
Equating the shift to the $n^{\text{th}}$ bright fringe position:
$\frac{D}{d} (\mu - 1)t = n \frac{\lambda D}{d} \implies (\mu - 1)t = n\lambda$
Step 3: Detailed Explanation:
Given parameters:
Refractive index $\mu = 1.56$
Fringe shift $n = 7$ (central fringe shifts to the 7th bright fringe position)
Wavelength $\lambda = 450 \text{ nm} = 450 \times 10^{-9} \text{ m}$
Use the derived equation:
\[ (\mu - 1)t = n\lambda \] \[ (1.56 - 1)t = 7 \times 450 \times 10^{-9} \] \[ 0.56 t = 3150 \times 10^{-9} \] Solve for $t$:
\[ t = \frac{3150 \times 10^{-9}}{0.56} \] To simplify the division, multiply numerator and denominator by 100:
\[ t = \frac{315000}{56} \times 10^{-9} \] Divide by 7:
\[ t = \frac{45000}{8} \times 10^{-9} \] Divide by 8:
\[ 45000 / 8 = 5625 \] \[ t = 5625 \times 10^{-9} \text{ m} \] The problem states thickness is $\alpha \times 10^{-9} \text{ m}$.
Thus, $\alpha = 5625$.
Step 4: Final Answer:
The value of $\alpha$ is 5625.