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Chapter 35: Q100P (page 1080)

A thin film suspended in air is 0.410 μmthick and is illuminated with white light incident perpendicularly on its surface. The index of refraction of the film is 1.50. At what wavelength will visible light that is reflected from the two surfaces of the film undergo fully constructive interference?

Short Answer

Expert verified

Thus, the wavelength of visible light is 492nm.

Step by step solution

01

According to the question.

The formula for the constructive interference:

2n2L=(m+12)λ

Here,λis wavelength of light in air/vacuum.

02

The wavelength of visible light. 

Use the above formula as follows:

λ=2n2Lm+12=21.50410 nmm+12=1230nmm+12

Where m=0,1,2,...The only value of mwhich, when substituted into equation above, would yield a wavelength that within the visible light range ism=1 .

Therefore, the wavelength is:

λ=1230 nm1+12=492 nm

Hence, the wavelength of visible light is492nm.

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