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:
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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 calculated using the formula:
\[
\beta = \frac{\lambda D}{d},
\]
where:
- \( \lambda \) denotes the wavelength of light in the medium,
- \( D \) represents the distance from the screen to the slits,
- \( d \) is the separation between the slits.
The wavelength of light when immersed in a liquid is determined by:
\[
\lambda' = \frac{\lambda}{n},
\]
with \( n = 1.44 \) being the refractive index of the liquid.
Upon substitution of the values, the wavelength in the liquid is found to be:
\[
\lambda' = \frac{690 \times 10^{-9}}{1.44} = 479.17 \, \text{nm}.
\]
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: \( 0.33 \, \text{mm} \).
