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The angular width of the central maximum in a single slit diffraction pattern is 60^0. The width of the slit is 1 μm. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance ?
(i.e. distance between the centres of each slit.)
 

  • Option 1)

    100 μm
     

  • Option 2)

    25 μm
     

  • Option 3)

    50 μm
     

  • Option 4)

    75 μm

 

Answers (2)

best_answer

As we learnt that

 

Fraunhofer Diffraction -

b\sin \theta = n\lambda
 

- wherein

Condition of nth minima.

b= slit width

\theta = angle of deviation

 

 semi\: angular\: width\: \theta =30^{o}

since\: bsin\theta =\lambda \Rightarrow \lambda =\frac{1}{2}\times 10^{-6}m

and \: fringe \: width\: \beta =\frac{\lambda D}{d}

\Rightarrow 10^{-2}=\frac{\frac{10^{-6}}{2}\times 0.50}{d}\: or\: d=25\mu m

 


Option 1)

100 μm
 

Option 2)

25 μm
 

Option 3)

50 μm
 

Option 4)

75 μm

Posted by

Avinash

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