1.

A double-slit experiment is immersed in a liquid of refractive index 1.33. It has slit separation of 1 mm and distance between the plane of slits and screen is 1.33 m. The slits are illuminated by a parallel bam of light whose wavelength in air is `6830 Å`. Then the fringe width isA. `6.3 xx 10^(-4) m`B. `8.3 xx 10^(-4) m`C. `6.3 xx 10^(-2) m`D. `6.3 xx 10^(-5) m`

Answer» Correct Answer - a
The experimental set-up is in a liquid, therefore the wavelength of light will change
`lambda_("liquid") = (lambda_("air"))/(mu) = (6300)/(1.33) = (6300 xx 10^(-10))/(1.33) m`
Fringe width,
`beta = lambda_("liquid") D/(d) = (lambda_("air") D)/(mu d) = (6300 xx 10^(-10))/(1.33) xx (1.33)/(10^(-3))`
`= 6.3 xx 10^(-4) m`


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