White light may be considered to be a mixture | vedclass

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(A) For a thin film of thickness $t$ and refractive index $\mu$,the condition for constructive interference (bright appearance) in reflected light is

Vedclass · Accepted Answer

$4308\ \mathring A, 5091\ \mathring A, 6222\ \mathring A$

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(A) For a thin film of thickness $t$ and refractive index $\mu$,the condition for constructive interference (bright appearance) in reflected light is given by $2\mu t \cos r = (2n - 1) \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$ and $r$ is the angle of refraction.Since the light is examined normally,$r = 0$,so $\cos r = 1$.The formula becomes $2\mu t = (2n - 1) \frac{\lambda}{2}$,which simplifies to $\lambda = \frac{4\mu t}{2n - 1}$.Given $\mu = 1.4$ and $t = 10,000\ \mathring A$,we have $\lambda = \frac{4 	imes 1.4 	imes 10,000}{2n - 1} = \frac{56,000}{2n - 1}\ \mathring A$.For $n = 1: \lambda = 56,000\ \mathring A$ (out of range).For $n = 2: \lambda = \frac{56,000}{3} \approx 18,666\ \mathring A$ (out of range).For $n = 3: \lambda = \frac{56,000}{5} = 11,200\ \mathring A$ (out of range).For $n = 4: \lambda = \frac{56,000}{7} = 8,000\ \mathring A$ (out of range).For $n = 5: \lambda = \frac{56,000}{9} \approx 6,222\ \mathring A$ (in range).For $n = 6: \lambda = \frac{56,000}{11} \approx 5,091\ \mathring A$ (in range).For $n = 7: \lambda = \frac{56,000}{13} \approx 4,308\ \mathring A$ (in range).Thus,the wavelengths are $4308\ \mathring A, 5091\ \mathring A, 6222\ \mathring A$.

White light may be considered to be a mixture of waves with $\lambda$ ranging between $3900\ \mathring A$ and $7800\ \mathring A$. An oil film of thickness $10,000\ \mathring A$ is examined normally by reflected light. If $\mu = 1.4$,then the film appears bright for

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