Q45P - University Physics with Modern Physics

Question

A thin uniform film of refractive index 1.750 is placed on a sheet of glass of refractive index 1.50. At room temperature 120.0°C2, this film is just thick enough for light with wavelength 582.4 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 170°C, you find that the film cancels reflected light with a wavelength of 588.5 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.)

Step-by-Step Solution

Answer

The coefficient of linear expansion of the film is $68\times {10}^{-6}{K}^{-1}$ .

Step 1: Linear Expresion

We start from $2t=\frac{m\lambda }{2n}$ for m = 1 in two cases and combine it with $∆t=a{t}_i∆T$

$$\begin{gathered} t_i=\frac{{\lambda }_{oi}}{2n}=\frac{582.4\times {10}^{-9}}{2\times 1.75}=166.4\times {10}^{-9} \\ {t}_i=\frac{{\lambda }_{of}}{2n}=\frac{582.4\times {10}^{-9}}{2\times 1.75}=168.14\times {10}^{-9} \\ ∆t=t_f-t_i=1.7\times {10}^{-9}=a{t}_i∆T \\ a=\frac{1.7\times {10}^{-9}}{166.4\times {10}^{-9}\times 150} \\ a=68\times {10}^{-9}{K}^{-1} \end{gathered}$$

Step 2: Conclusion

Hence, the coefficient of linear expansion of the film is $68\times {10}^{-6}{K}^{-1}$