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 Interference pattern is observed at ‘P’ due to superimposition of two rays coming out 
from a source ‘S’ as shown in the figure.The value of ‘l’ for which maxima is
obtained at ‘P’ is : (R is perfect reflecting surface) :    

  • Option 1)

    l= \frac{2n\lambda }{\sqrt{3}-1}

  • Option 2)

    l= \frac{\left ( 2n-1 \right )\lambda }{2\left ( \sqrt{3}-1 \right )}

  • Option 3)

    l= \frac{\left ( 2n-1 \right )\lambda\sqrt{3} }{4\left ( 2-\sqrt{3} \right )}

  • Option 4)

    l= \frac{\left ( 2n-1 \right )\lambda }{\sqrt{3}-1}

 

Answers (1)

As we have learned

Lloyd's mirror Experiment -

S_{2}P-S_{1}P= n\lambda

S_{2}P-S_{1}P= \left ( n+1/2 \right )\lambda

- wherein

minima

maxima

 

 For maxima at P 

\Delta x= (n-1/2 )\lambda

x \cos 30 \degree = l \\ \Rightarrow x = \frac{2l }{\sqrt 3 }

\Delta x = \frac{2l}{\sqrt 3}+ \frac{2l}{\sqrt 3 } - 2l = (n- 1/2 )\lambda

Or 

\frac{4l}{\sqrt3}-2l = (n-1/2 )\lambda

or 

l = \frac{(2n-1)\lambda \sqrt 3 }{4(2-\sqrt 3 )}

 

 

 

 

 

 

 

 

 

 

 

 

 


Option 1)

l= \frac{2n\lambda }{\sqrt{3}-1}

Option 2)

l= \frac{\left ( 2n-1 \right )\lambda }{2\left ( \sqrt{3}-1 \right )}

Option 3)

l= \frac{\left ( 2n-1 \right )\lambda\sqrt{3} }{4\left ( 2-\sqrt{3} \right )}

Option 4)

l= \frac{\left ( 2n-1 \right )\lambda }{\sqrt{3}-1}

Posted by

Vakul

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