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In a single slit diffraction experiment ...

In a single slit diffraction experiment first minima for `lambda_1 = 660nm` coincides with first maxima for wavelength `lambda_2`. Calculate the value of `lambda_2`.

Text Solution

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For minima in diffraction pattern `dsintheta=nlamda`
for first minima `dsintheta_(1)=(1)lamda_(1)impliesintheta_(1)=(lamda_(1))/(d)`
for first maxima `dsintheta_(2)=(3)/(2)lamda_(2)impliessintheta_(2)=(3lamda_(2))/(2d)`
The two will coincide if `theta_(1)=theta_(2)" or "sintheta_(1)=sintheta_(2)`
`therefore(lamda_(1))/(d)=(3lamda_(2))/(2d)implieslamda_(2)=(2)/(3)lamda_(1)=(2)/(3)xx660nm=440nm`
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