Question

In Young's double slit experiment, the angular width of a fringe is found to be \(0.2^{\circ}\) on a screen placed 1m away. The wavelength of light used in 600 nm. If the entire apparatus is immersed in water of refractive index 4/3. the angular width of the fringe will be

• A. \(0.10^{\circ}\)
• B. \(0.15^{\circ}\)
• C. \(0.20^{\circ}\)
• D. \(0.25^{\circ}\)

Hint:

The wavelength in water is the air wavelength divided by the refractive index, and angular width is proportional to wavelength.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The angular fringe width is \(\theta = \dfrac{\lambda}{d}\). When the apparatus goes into a medium of index \(\mu\), the wavelength becomes \(\lambda/\mu\), while \(d\) stays the same.

Step 2: Calculate.
\[ \theta_{water} = \frac{\theta_{air}}{\mu} = \frac{0.2^\circ}{4/3} = 0.2^\circ\times\frac{3}{4} = 0.15^\circ \]

Step 3: Check the options.
\(0.10^\circ\) would need \(\mu = 2\). \(0.20^\circ\) means no change, which would be true only in air. \(0.25^\circ\) is bigger and would mean the fringes widened.

Final Answer:
The angular width becomes \(0.15^\circ\), option (B). \[ \boxed{0.15^\circ} \]