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A long narrow horizontal slit lies 1 mm ...

A long narrow horizontal slit lies `1 mm` above a plane mirror. The interference pattern produced by the slit and its and its image is viewed on a screen `sum` distance `1m` from the slit. The wavelength of light is `600 nm`. Then the distance of the first maxima above the mirror is equal to `(d lt lt D)` :

A

`0.30 mm`

B

`0.15 mm`

C

`60 mm`

D

`7.5 mm`

Text Solution

Verified by Experts

The correct Answer is:
B
Point `O` is a minima Hence the maxima will be at `y=(beta)/(2)` from `O.`
`rArr y=(lambdaD)/(2(2d))=(600xx10^(-9)xx1)/(4xx1xx10^(-3)) =0.15 mm.`
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