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  • Answer please! 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)

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 :

  • Option 1)

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

  • Option 2)

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

  • Option 3)

    a= \sqrt{\lambda L}\: \: \: and\: \: b_{min}= \sqrt{4\lambda L}

  • Option 4)

    a=\frac{\lambda ^{2}}{L}\: \: and \: \: b_{min}=\sqrt{4\lambda L}

 

Answers (1)

best_answer

As we have learnt,

 

Fraunhofer Diffraction -

b\sin \theta = n\lambda
 

- wherein

Condition of nth minima.

b= slit width

\theta = angle of deviation

 

 

Considering the gemetrical spreading of light alone

O'A' = OA = Q

Freinel's displacement 

L = \frac{\lambda^2}{2Q}\;\;\;-(1)

Considering the Freinel's class of diffraction

Angle of diffraction is given by the relation a\theta = \lambda \;\;\; -(2)

From fig: \theta = \frac{a}{L}

a\cdot \frac{a}{L} = \lambda \Rightarrow a = \sqrt{\lambda L}

From figure b_{min} = 2 a = \sqrt{4\lambda L} 

 


Option 1)

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

Option 2)

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

Option 3)

a= \sqrt{\lambda L}\: \: \: and\: \: b_{min}= \sqrt{4\lambda L}

Option 4)

a=\frac{\lambda ^{2}}{L}\: \: and \: \: b_{min}=\sqrt{4\lambda L}

Posted by

Avinash

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