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Light of wavelength 550 nm falls normally on a slit of width 22.0×10^{-5} cm. The angular position of the second minima
from the central maximum will be (in radians) :

  • Option 1)

    \frac{\pi }{12}

     

     

     

  • Option 2)

    \frac{\pi }{8}

  • Option 3)

    \frac{\pi }{6}

  • Option 4)

    \frac{\pi }{4}

 

Answers (2)

best_answer

As we learned

 

Fraunhofer Diffraction -

b\sin \theta = n\lambda
 

- wherein

Condition of nth minima.

b= slit width

\theta = angle of deviation

 

 for second minima b \sin \theta =2\lambda

 

\Rightarrow \sin \theta =\frac{2\lambda}{b}= \frac{2\times 550\times 10^{-9}}{22\times 10^{-7}}= 0.50

\theta =\frac{\pi }{6}

 


Option 1)

\frac{\pi }{12}

 

 

 

Option 2)

\frac{\pi }{8}

Option 3)

\frac{\pi }{6}

Option 4)

\frac{\pi }{4}

Posted by

Aadil

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