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Light of wavelength 6328 Å is incident n...

Light of wavelength `6328 Å` is incident normally on a slit having a width of `0.2 mm`. The distance of the screen from the slit is `0.9 m`. The angular width of the central maximum is

A

2.4 mm

B

1.2 mm

C

0.6 mm

D

4.8 mm

Text Solution

Verified by Experts

The correct Answer is:
B
The angular position of m= 1 minima are given by
`sintheta=pm(lambda)/(d)`
`therefore`   From the adjacent figure, `sintheta=(y)/sqrt((y^(2)+D^(2)))`
If `Deltay` is the distance between the two m = 1 minima, the values of y that correspond to m 1 minima are
`y=pm(1//2)Deltay`
`therefore` `((1)/(2)Deltay)/sqrt(((1)/(2)Deltay)^(2)+D^(2))=(lambda)/(D)`  `rArr` `Deltay=(2D)/sqrt(d^(2)-lambda^(2))`
If `dimplies>lambda` `sqrt(d^(2)-lambda^(2))congd`
`therefore`   `Deltay=(2Dlambda)/(d)=(2xx0.60xx600xx10^(_9))/(0.6xx10^(-3))=12xx10^(-4)=1.2mm`
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