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Chapter 17: Q. 67 (page 486)

Scientists are testing a transparent material whose index of refraction for visible light varies with wavelength as n = 30.0 nm1/2 /l1/2, where l is in nm. If a 295-nm-thick coating is placed on glass 1n = 1.502 for what visible wavelengths will the reflected light have maximum constructive interference?

Short Answer

Expert verified

Here,λc=679.18nmλc=427.85m

Step by step solution

01

Understanding the constructive interference 

Constructive interference occurs when the maxima of two waves add together so that the amplitude of the resulting wave is equal to the sum of the individual amplitudes

02

Processing of the equation 

Here,d=295m

Thus, role="math" localid="1649154544207" n=30nm1/2λ1/2

The wavelength is λc=2ndmm=1,2,3...

The n= n=30nm1/2λ1/2the equation is

role="math" localid="1649154803634" λc=2dm·30nm1/2λ1/2λc3/2=d60nm1/2mλc=d2·3600nmm3

Thus, the solution to the equation is 5

for m=1 , role="math" localid="1649154869717" λc=2952·3600nmm3λc=679.18nm

for m=2, role="math" localid="1649155019072" λc=2952·3600nm23λc=679.18nm

For m=3, λc=2952·3600nm33λ=326.51nm

Thus, the interference range in 360 to 700 nanometers is 679.18 nm and 427.89 nm.

Thus,role="math" localid="1649155157302" λc=679.18nmλc=427.85nm

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