Question

In a Young's double-slit experiment with light of wavelength 520 nm, the fringe width is 0.52 mm. What will be the fringe width if the entire apparatus is immersed in water (refractive index \(\dfrac{4}{3}\))?

Hint:

In water the wavelength becomes lambda / n, so the fringe width also becomes beta / n.

Answer 1

Step 1: Write what changes in water:
Fringe width is \(\beta = \lambda D/d\). The apparatus is not changed, so \(D\) and \(d\) stay the same. Only the wavelength changes, since light slows down in water.
In a medium of index \(n\), the wavelength becomes \(\lambda_w = \lambda / n\). The frequency stays the same.

Step 2: Find the new wavelength:
\[ \lambda_w = \frac{520}{4/3} = 520 \times \frac{3}{4} = 390 \text{ nm} \]

Step 3: Find the new fringe width:
\(\beta_w / \beta = \lambda_w / \lambda = 1/n\).
\[ \beta_w = \frac{\beta}{n} = \frac{0.52}{4/3} = 0.52 \times \frac{3}{4} = 0.39 \text{ mm} \]

Final Answer:
The fringe width in water is 0.39 mm. It is smaller than in air. \[ \boxed{\beta_w = 0.39 \text{ mm}} \]