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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. (a) `6.3xx10^-4m`B. (b) `8.3xx10^-4m`C. (c) `6.3xx10^-2m`D. (d) `6.3xx10^-5m` |
Answer» Correct Answer - A The experimental setup is in a liquid, therefore the wavelength of light will change. `lambda_(liquid)=(lambda_(air))/(mu)=(6300)/(1.33)` `=(6300xx10^(-10))/(1.33)m` Fringe width `beta=(lambda_(liquid)D)/(d)=(lambda_(air)D)/(mud)=(6300xx10^(-10))/(1.33)xx(1.33)/(10^-3)` `=6.32xx10^-4m` |
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