Initially interference is observed with the entire experimental set up inside a chamber filled with air, Now the chamber is evacuated. With the same source of light used, a careful observer will find that.

The interference pattern is almost absent as it is very much diffused

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There is no change in the interference pattern

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The fringe width is slightly decreased

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The fringe width is slightly increased

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Solution

The correct option is D The fringe width is slightly increased
The refractive index of vacuum is less than that of air. The fringe width will vary as follows

$β_{a i r} = \frac{β_{v a c u u m}}{μ_{a i r}}$
And $μ_{a i r}$ is greater than 1. So, fringe width will slightly increase.

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Q. In a certain double slit experimental arrangement interference fringes of width

1.0

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each are observed when light of wavelength

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˚ A

is used. Keeping the set up unaltered, if the source is replaced by another source of wavelength

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˚ A

, the fringe width will be

In a certain double slit experimental arrangement interference fringes of width 1.0 mm each are observed when light of wavelength 5000 Å is used. Keeping the set up unaltered, if the source is replaced by another source of wavelength 6000 Å, the fringe width will be

[CPMT 1988]

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Initially interference is observed with the entire experimental set up inside a chamber filled with air, Now the chamber is evacuated. With the same source of light used, a careful observer will find that.

The correct option is D The fringe width is slightly increasedThe refractive index of vacuum is less than that of air. The fringe width will vary as follows βair=βvacuumμair And μairis greater than 1. So, fringe width will slightly increase.

The correct option is D The fringe width is slightly increased
The refractive index of vacuum is less than that of air. The fringe width will vary as follows

$β_{a i r} = \frac{β_{v a c u u m}}{μ_{a i r}}$
And $μ_{a i r}$ is greater than 1. So, fringe width will slightly increase.