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 ( 22.2 ∘C), this film is just thick enough for light with a wavelength 583.5 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 174 ∘C, you find that the film cancels reflected light with a wavelength 587.5 nm .

Answer 1

`\alpha = 4.742*10^(-5)/ \°C`

Step-by-step explanation:

`n= 1.750`

`\lambda = 583.5nm`

In this way we understand that the condition for destructive interference is

`2t = (m \lamba)/(n)`

The smallest non zero thickness is,

`t= (\lambda)/(2n)`

At
`22.2\° C`

`t_0 = (583.5nm)/(2(1.750))`

`t_0 = 166.7nm`

At
`174\° C`

`t= (587.5nm)/(2(1.750))`

`t= 167.9nm`

`t= t_0 (1+\alpha \Delta T)`

The coefficient of linear expansion is

`\alpha = (t-t_0)/(t_0 \Delta T)`

`\alpha = (167.9nm-166.7)/(166.7(174\° - 22.2\°))`

`\alpha = 0.00004742/ \°CC`

`\alpha = 4.742*10^(-5)/ \°C`