A thin uniform film of refractive index 1.75 is placed on a sheet of glass of refractive index 1.5. At room temperature `(20^(@) C)` this film is just thick enoungh for light with wavelength 600 nm reflected off the top of the film to be canceled by light reflected from the top the glass. After the glass is placed in on oven and slowly heated to `170^(@) C`, the film conceals reflected light with wavelength 606 nm. The coefficient of linear expansion of the film is (ignore any changes in the refractive index of the film due to the temperature change)

Options

A. `3.3 xx 10^(-5).^@C^(-1)`
B. `6.6 xx 10^(-5).^@C^(-1)`
C. `9.9 xx 10^(-5).^@C^(-1)`
D. `2.2 xx 10^(-5).^@C^(-1)`

Answer

b

Explanation

For destructive reflection :
At `theta_(1) = 20^(@) C, (2mu_(1))/(mu_(g)) t_(1) = n lambda_(1)`
At `theta_(2) = 170^(@), (2 mu_(1))/(mu_(g)) = n lambda_(2)`
`(t_(2))/(t_(1)) = (lambda_(2))/(lambda_(1))`
`(t_(1) [1 + alpha (theta_(2) - theta_(1))])/(t_(1)) = (lambda_(2))/(lambda_(1))`
(`alpha` is the coefficient of linear expansion of the film)
`implies [1 + alpha (170 - 20)] = (606)/(600)`
or `alpha = (60)/(600 xx 150) = 6.6 xx 10^(-5).^@C^(-1)`