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The angular deviation of 5th order dark ...

The angular deviation of 5th order dark fringe is `12^(@)` in a single slit experiment. If the width of the slit is 9 `mu`m then the wavelength of the incident light is

A

4862 Å

B

5892 Å

C

6022 Å

D

3768 Å

Text Solution

Verified by Experts

The correct Answer is:
D
Given angular deviation of 5yh order dark fringe, `theta 12^(@)` ,
Width of the slit, d = 9 `mu` m
Now, angular deviation of nth fringe = n .`(lambda)/(d)`
` therefore 12 xx (pi)/(180) = 5 xx (lambda)/(9 xx 10^(-6))`
or `" " lambda = (12 xx pi)/( 180) xx (9 xx 10^(-6))/(5) `m
or ` " " lambda = 3769 `Å
So, the wave length of incident light is `lambda` = 3768 Å.
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