Question

A double slit experiment is immersed in water of refractive index \( 1.33 \). The slit separation is \( 1\,\text{mm} \), distance between slit and screen is \( 1.33\,\text{m} \). The slits are illuminated by light of wavelength \( 6300\,\text{\AA} \). The fringe width is

• A. \( 6.9 \times 10^{-4} \, \text{m} \)
• B. \( 6.3 \times 10^{-4} \, \text{m} \)
• C. \( 5.8 \times 10^{-4} \, \text{m} \)
• D. \( 8.6 \times 10^{-4} \, \text{m} \)

Hint:

When a double slit experiment is performed in a medium, divide the wavelength by refractive index.

Answer 1

The Correct Option is B

Step 1: Write formula for fringe width.
\[ \beta = \frac{\lambda D}{d} \]
Step 2: Modify wavelength in medium.
In a medium of refractive index \( \mu \), \[ \lambda' = \frac{\lambda}{\mu} \]
Step 3: Substitute given values.
\[ \lambda = 6300\,\text{\AA} = 6.3 \times 10^{-7}\,\text{m} \] \[ \mu = 1.33, \quad D = 1.33\,\text{m}, \quad d = 1 \times 10^{-3}\,\text{m} \]
Step 4: Calculate fringe width.
\[ \beta = \frac{\lambda D}{\mu d} = \frac{6.3 \times 10^{-7} \times 1.33}{1.33 \times 10^{-3}} = 6.3 \times 10^{-4}\,\text{m} \]
Step 5: Conclusion.
The fringe width is \( 6.3 \times 10^{-4}\,\text{m} \).