A soap film of 5*10^-5 cm thick is viewed at an angle of 35 degree to normal find the wavelenth of line in the

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.