Question

A Young's double slit apparatus has slits separated by \(0.28\,\text{mm}\) and a screen \(48\,\text{cm}\) away from the slits. The whole apparatus is immersed in water, and the slits are illuminated by red light \(\lambda=700\,\text{nm}\) in vacuum. The fringe width of the pattern formed on the screen is:

• A. \(0.90\,\text{mm}\)
• B. \(0.60\,\text{mm}\)
• C. \(0.80\,\text{mm}\)
• D. \(0.40\,\text{mm}\)

Hint:

In water use \(\lambda_w=\lambda_{vac}/n\) with \(n=\tfrac{4}{3}\), then \(\beta=\lambda_w D/d\).

Answer 1

The Correct Option is A

Step 1: The fringe width in a double slit experiment is \[\beta=\frac{\lambda D}{d}\] where \(\lambda\) is the wavelength in the medium, \(D\) is the slit-to-screen distance, and \(d\) is the slit separation.

Step 2: Inside water the wavelength shrinks by the refractive index \(n=\tfrac{4}{3}\): \[\lambda_{w}=\frac{\lambda_{\text{vac}}}{n}=\frac{700\,\text{nm}}{4/3}=700\times\frac{3}{4}=525\,\text{nm}\]
Step 3: Convert data to SI: \(\lambda_w=525\times10^{-9}\,\text{m}\), \(D=0.48\,\text{m}\), \(d=0.28\times10^{-3}\,\text{m}\). \[\beta=\frac{525\times10^{-9}\times0.48}{0.28\times10^{-3}}\]
Step 4: Numerator \(=2.52\times10^{-7}\). Divide: \[\beta=\frac{2.52\times10^{-7}}{2.8\times10^{-4}}=9.0\times10^{-4}\,\text{m}=0.90\,\text{mm}\] \[\boxed{\beta=0.90\,\text{mm}}\]