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The box of a pin hole camera, of length ...

The box of a pin hole camera, of length L, has a hole of radius a . It is assumed that when the hole is illuminated by a parallel beam of light of wavelength `lamda` the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say b_(min)) when:

A

a = `lambda^(2)/L` and b_(min) = `((2lambda^(2))/L)`

B

a = `sqrt(lambda L)` and b_(min) = `((2lambda^(2))/L)`

C

a = `sqrt(lambda L)` and b_(min) = `sqrt(4 lambda L)`

D

a = `lambda^(2)/L ` and b_(min) =`sqrt(4 lambda L)`

Text Solution

Verified by Experts

The correct Answer is:
3

sin`theta = lambda/a `, B = 2a + `(a L lambda)/a` ….........(1)
`B/a` = 0 `implies 1 - (L lambda)/(a^(2)) = 0 implies a sqrt( lambda L)` …...(ii)
`B_(min) = 2 sqrt(lambda L) + 2 sqrt(lambda L )` [By substituting for a from
(ii)in(i)]
= 4`sqrt(lambda L)`
`:. ` The radius of the spot = `1/2 4 sqrt(lambda L) = sqrt(4 lambda L )`
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