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A soap film of \(5\times10^{-5 }\) cm thick is viewed at an angle of 35 degree to normal find the wavelenth of line in the visible spectrum which will be acts at from the reflected light μ{mu}=1.33
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Given: angle of incident =i=35o 

thickness=t=5*10-5cm 

refractive index=μ= 1.33

Solution :

\(\mu=\frac{\sin i}{\sin r}\)

\(\sin r=\frac{\sin i}{\mu}\)

 \(\sin r=0.43\) 

Therefore, r = 25.46 

\(\cos r = 0.902\)  

Condition for darkness is \(2μ t \cos r = nλ \)

 Substitute values of n as 1,2,3,4……… we will get corresponding wavelengths .those wavelengths which fall in the visible spectra will remain absent.  

For n=1 \(\lambda_1=2\times0.902\times1.33\times5\times10^{-5}\)  = 11.9966 \(\times10^{-5}\)  cm =11996.6 Å 

For n=2 \(\lambda_2=\frac{2\times0.902\times1.33\times5\times10^{-5}}{2}\)  = 5.9983\(\times 10^{-5 }\) cm =5998.3 Å

For n=3 \(\lambda_3=\frac{2\times0.902\times1.33\times5\times10^{-5}} {3}\) =3.998867 \(\times 10^{-5}\) cm =3998.867 Å 

Hence ,  the wavelengths in the reflected light 5998.3 Å is in the visible region of spectrum.

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