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Question

Physics Question on Youngs double slit experiment

The box of a pin hole camera, of length LL, has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength λ\lambda 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 bminb_{min}) when :

A

a=λ2La = \frac{\lambda^{2}}{L} and bmin=(2λ2L)b_{min } = \left(\frac{2 \lambda^{2}}{L}\right)

B

a=λLa = \sqrt{\lambda L} and bmin=(2λ2L)b_{min } = \left(\frac{2 \lambda^{2}}{L}\right)

C

a=λLa = \sqrt{\lambda L} and bmin=4λL b_{min } = \sqrt{4 \lambda L}

D

a=λ2La = \frac{\lambda^2}{ L} and bmin=4λL b_{min } = \sqrt{4 \lambda L}

Answer

a=λLa = \sqrt{\lambda L} and bmin=4λL b_{min } = \sqrt{4 \lambda L}

Explanation

Solution

sinθ=λa\sin\theta = \frac{\lambda}{a}

B=2a+2LλaB = 2a + \frac{2L\lambda}{a} .....(i)
Ba=0\frac{\partial B}{\partial a} = 0
1Lλa2=0\Rightarrow 1- \frac{L \lambda}{a^{2}} = 0
a=λL\Rightarrow a = \sqrt{\lambda L} ....(ii)
Bmin=2λL+2λLB_{min } = 2\sqrt{\lambda L} + 2 \sqrt{\lambda L} \,\,\,\, [By substituting for a from (ii) in (i)]
=4λL= 4 \sqrt{\lambda L}
\therefore The radius of the spot =124λL=4λL= \frac{1}{2 } 4 \sqrt{\lambda L} = \sqrt{4\lambda L}