Question

Distinguish between interference and diffraction of light. A double slit arrangement produces interference fringes for sodium light of wavelength 589 nm, that are 0.20 degree apart. What is the angular fringe separation if the entire arrangement is immersed in water? (R.I. of water = 1.33)

Hint:

When an optical experiment is moved from air into a medium like water, the only thing that changes is the wavelength of light (\(\lambda_{new} = \lambda_{air}/n\)). This makes the fringes closer together.

Answer 1

Distinction between Interference and Diffraction: 

Calculation: The angular fringe separation (\(\theta\)) in a double-slit experiment is given by: \[ \theta = \frac{\lambda}{d} \] where \(\lambda\) is the wavelength of light and \(d\) is the distance between the slits.
In air:
Wavelength, \(\lambda_{air} = 589 \, nm\)
Angular separation, \(\theta_{air} = 0.20^{\circ}\)
So, \(0.20^{\circ} = \frac{\lambda_{air}}{d}\).
When the apparatus is immersed in water:
The slit separation \(d\) remains the same.
The wavelength of light changes. The new wavelength in water is \(\lambda_{water}\).
Refractive index of water, \(n_w = 1.33\).
The relationship between wavelengths is \(\lambda_{water} = \frac{\lambda_{air}}{n_w}\).
The new angular fringe separation in water (\(\theta_{water}\)) will be: \[ \theta_{water} = \frac{\lambda_{water}}{d} = \frac{(\lambda_{air}/n_w)}{d} = \frac{1}{n_w} \left(\frac{\lambda_{air}}{d}\right) \] Since we know \(\frac{\lambda_{air}}{d} = \theta_{air}\), we can substitute this into the equation: \[ \theta_{water} = \frac{\theta_{air}}{n_w} \] \[ \theta_{water} = \frac{0.20^{\circ}}{1.33} \approx 0.1503^{\circ} \] The new angular fringe separation is approximately 0.15\(^{\circ}\).