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Light of wavelength lambda=5000Å falls n...

Light of wavelength `lambda=5000Å` falls normally on a narrow slit. A screen is placed at a distance of `1m` from the slit and perpendicular to the direction of light. The first minima of the diffraction pattern is situated at `5mm` from the centre of central maximum. The width of the slit is

A

(a) `0.1mm`

B

(b) `1.0mm`

C

(c) `0.5mm`

D

(d) `0.2mm`

Text Solution

Verified by Experts

The correct Answer is:
A
Position of nth minima `x_n=(nlambdaD)/(d)`
`implies5xx10^-3=(1xx500xx10^(-10)xx1)/(d)`
`impliesd=10^-4m=0.1mm`.
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