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 λ, 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 bmin) when

(B) a = \frac{{{\lambda ^2}}}{L} and b_{min}=\sqrt {4\lambda L}

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

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

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JEE Mains · Physics · STD 12 - Wave optics

JEE Mains 2016
Hard
MCQ

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 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

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Answer & Solution

Correct Answer

(A) \(a = \sqrt {\lambda L} \) and \(b_{min}\)=\(\sqrt {4\lambda L} \)

Step-by-step Solution

Detailed explanation

Given geometrical spread \(=\,a\)

Diffraction spread \(=\frac{\lambda}{\mathrm{a}} \times \mathrm{L}=\frac{\lambda \mathrm{L}}{\mathrm{a}}\)

The sum \(b=a+\frac{\lambda L}{a}\)

For \(b\) to be minimum \(\frac{{{\text{db}}}}{{{\text{da}}}} = 0\) \(\frac{{\text{d}}}{{{\text{da}}}}\left( {{\text{a}} + \frac{{\lambda {\text{L}}}}{{\text{a}}}} \right) = 0\)

\(a =\sqrt{\lambda L}\)

\(b\, \min =\sqrt{\lambda L}+\sqrt{\lambda L}=2 \sqrt{\lambda L}=\sqrt{4 \lambda L}\)

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