Question

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

• A. \(4.9\times 10^{-4}\) m
• B. \(6.3\times 10^{-4}\) m
• C. \(8.6\times 10^{-4}\) m
• D. \(5.8\times 10^{-4}\) m

Hint:

Fringe width = lambda D / d with the wavelength in the medium lambda / n.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In a medium of index n, the wavelength becomes \(\lambda_m = \lambda/n\), so the fringe width is \(\beta = \dfrac{\lambda D}{n\,d}\). The wavelength given (6300 angstrom) is the one in air.

Step 2: Compute
\[ \beta = \frac{6300\times10^{-10}\times 1.33}{1.33\times 1\times10^{-3}} = 6.3\times10^{-4}\text{ m} \]
The 1.33 in D and the 1.33 in n cancel. Option (A) would result if the water factor were used twice.

Final Answer:
The fringe width is \(6.3\times10^{-4}\) m, option (B). \[ \boxed{6.3\times10^{-4}\text{ m}} \]