A thin uniform film of refractive index 1.750 is placed on a sheet of glass with

Question

A thin uniform film of refractive index 1.750 is placed on a sheet of glass with a refractive index 1.50. At room temperature (24.0 degree C), this film is just thick enough for light with a wavelength 581.8 nm reflected off the top of the film to be canceled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to 169 degree C, you find that the film cancels reflected light with a wavelength 589.2 nm. What is the coefficient of linear expansion of the film? (Ignore any changes in the refractive index of the film due to the temperature change.) Express your answer using two significant figures.

Explanation / Answer

According to the given question,

Change in Temperature = 169-24 = 145 'C

Change in thickness = Change in the wavelength of light that gets refracted. = (589.2 - 581.8) nm = 7.4 nm = 7.4 * 10-9 m

So, from linear thermal expansion formula,

Change in length = Original L * Alpha * Change in T

Alpha = Change in length /(Original L* Change in T) =

By putting values, we get Alpha = 8.77 * 10-5 'C

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