Question

Measurements are made of the intensity distribution within the central bright fringe in a Young's interference pattern. At a particular value of y, it is found that I/Imax= 0.810 when 600-nm light is used. What wavelength of light should be used to reduce the relative intensity at the same location to 64.0% of the maximum untensity?

Answer 1

As per the question,

$\frac{I}{I_{max}}=0.810$

$\lambda_1=600nm =600\times 10^{-9}m$

$\lambda_2=?$

relative intensity at the same location $=64.0\%$

$I=I_{max}\cos^2({\frac{\pi yd}{\lambda_1 L})}$

$\Rightarrow \frac{\pi yd}{ L}=\lambda_1\cos^{-1}(\frac{I}{I_{max}})^{1/2}$

$\Rightarrow \frac{\pi y d}{L}=600 \cos^{-1}(0.810)^{1/2}$

$\Rightarrow \frac{\pi y d}{L}=271 nm$

Now,

$\frac{I}{I_{max}}=0.64$

$\lambda_2=\frac{\frac{\pi y d}{L}}{\cos^{-1}(\frac{I}{I_{max}})^{1/2}}=38.14nm$