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  • Help me solve this - Consider a Young's double slit experiment as shown in figure. What should be the slit sepration d in - Optics - JEE Main

Consider a Young's double slit experiment as shown in figure. What should be the slit sepration d in terms of wavelength \lambda such that the first minima occurs directly in front of the slit (S1)?

 

  • Option 1)

    \frac{\lambda }{2\left ( 5-\sqrt{2} \right )}

  • Option 2)

    \frac{\lambda }{\left (\sqrt{5}-2 \right )}

  • Option 3)

    \frac{\lambda }{2\left (\sqrt{5}-2 \right )}

  • Option 4)

     

    \frac{\lambda }{\left ( 5-\sqrt{2} \right )}

Answers (1)

best_answer

 

Minimum amplitude & Intensity -

\theta = \left ( 2n+1 \right )\pi
 

- wherein

A_{min }= A_{1}-A_{2}

I_{min }= \left ( \sqrt{I_{1}}-\sqrt{I_{2}} \right )^{2}

 

 

Young Double Slit Experiment -

- wherein

y=\Delta x\cdot \left ( \frac{D}{d} \right )

y= Distance of a point on screen from central maxima

\Delta x= Path difference at that point

 

x1 = 2d,   x_{2}= \sqrt{5}d

\Delta x = x_{2}-x_{1} = d\left ( \sqrt{5}-2 \right )

d=\frac{\lambda }{2\left ( \sqrt{5}-2 \right )}

 


Option 1)

\frac{\lambda }{2\left ( 5-\sqrt{2} \right )}

Option 2)

\frac{\lambda }{\left (\sqrt{5}-2 \right )}

Option 3)

\frac{\lambda }{2\left (\sqrt{5}-2 \right )}

Option 4)

 

\frac{\lambda }{\left ( 5-\sqrt{2} \right )}

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