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 `lambda`, the spread of the spot (obtained on the opposite wall of camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would have its minimum size (say `b_(min)`) when.

Options

A. `a = (lambda^(2))/(L) and b_(min) = (2lambda^(2))/(L)`
B. `a = sqrt(lambdaL) and b_(min) = (2lambda^(2))/(L)`
C. `a = sqrt(lambdaL) and b_(min) = sqrt(4lambdaL)`
D. `a = (lambda^(2))/(L) and b_(min) = sqrt(4lambdaL)`

Explanation

`sin theta = (lambda)/(a)`
`tan theta = (CE)/(AE)`
Since `theta` is very small, so ten`theta approx sin theta`
`therefore " " sin theta = (CE)/(AE) = or, (lambda)/(a) = (CE)/(L) or, CE = (Llambda)/(a)`
Hence diameter of the spot,
` B = EF + CE + FD = 2a + (2Llambda)/(a)`
The diameter of the spot will be minimum when,
`(dB)/(da) = 0`
or, `2 - (2Llambda)/(a^(2))` = 0 or,` (Llambda)/(a^(2)) = 1 or, a = sqrt(Llambda)`
`therefore " " B_(min) = 2sqrt(L lambda) + 2 sqrt(L lambda) = 4 sqrt(L lambda)`
Hence radius of the spot,
`b_(min) = (1)/(2) xx 4 sqrt(L lambda) = 2sqrt(L lambda) = sqrt(4L lambda)`
The option (C) is correct.