Young's double slit interference apparatus is immersed in a liquid of refractive index 1.44. It has slit separation of 1.5 mm. The slits are illuminated by a parallel beam of light whose wavelength in air is 690 nm. The fringe-width on a screen placed behind the plane of slits at a distance of 0.72 m, will be:
Show Hint
In a Young's double slit experiment, the fringe width depends on the wavelength of light, the distance between the slits, and the distance between the screen and the slits. Remember to adjust the wavelength according to the refractive index when in a medium other than air.
The fringe width \( \beta \) in a Young's double slit experiment is determined by the formula:
\[
\beta = \frac{\lambda D}{d},
\]
where \( \lambda \) represents the wavelength of light in the medium, \( D \) is the distance from the screen to the slits, and \( d \) is the separation between the slits.
The wavelength of light within the liquid is calculated as:
\[
\lambda' = \frac{\lambda}{n},
\]
with \( n = 1.44 \) being the refractive index of the liquid.
Upon substitution of values:
\[
\lambda' = \frac{690 \times 10^{-9}}{1.44} = 479.17 \, \text{nm}.
\]
Consequently, the fringe width in the liquid is:
\[
\beta = \frac{479.17 \times 10^{-9} \times 0.72}{1.5 \times 10^{-3}} = 0.33 \, \text{mm}.
\]
Final Answer: \( 0.33 \, \text{mm} \).
