Question

A thin uniform film of refractive index $1.75$ is placed on a sheet of glass of refractive index $1.5$. At room temperature $\left(20{}^{\circ }\text{C}\right)$, this film is just thick enough for light with wavelength $600\text{nm}$ reflected off the top of the film to be canceled by light reflected from top of the glass. After the glass is placed in an oven and slowly heated to $170{}^{\circ }\text{C}$, the film cancels reflected light of wavelength $606\text{nm}$. The coefficient of linear expansion of the film is (Ignore any changes in the refractive index of the film due to the temperature change.)

• A. $(3.3\times {10}^{-5}){}^{\circ }{\text{C}}^{-1}$
• B. $(6.6\times {10}^{-5}){}^{\circ }{\text{C}}^{-1}$
• C. $(9.9\times {10}^{-5}){}^{\circ }{\text{C}}^{-1}$
• D. $(2.2\times {10}^{-5}){}^{\circ }{\text{C}}^{-1}$

Answer 1

The correct option is B $(6.6\times {10}^{-5}){}^{\circ }{\text{C}}^{-1}$

According to the question,

For destructive reflection,

At temperature, ${\theta }_1={20}^{\circ }\text{C}$

$\frac{2{\mu }_1}{{\mu }_g}{t}_1=n{\lambda }_1......\left(1\right)$

At temperature, ${\theta }_2={170}^{\circ }\text{C}$

$\frac{2{\mu }_1}{{\mu }_g}{t}_2=n{\lambda }_2.........\left(2\right)$

From equations $\left(1\right)$ and $\left(2\right)$, we get

$\frac{t_2}{t_1}=\frac{{\lambda }_2}{{\lambda }_1}$

$\frac{t_1[1+\alpha ({\theta }_2-{\theta }_1\left)\right]}{t_1}=\frac{{\lambda }_2}{{\lambda }_1}$

Where, $\alpha$ is the coefficient of linear expansion of the film

$\Rightarrow [1+\alpha (170-20\left)\right]=\frac{606}{600}$

$\therefore \alpha =\frac{6}{600\times 150}=(6.6\times {10}^{-5}){}^{\circ }{\text{C}}^{-1}$

Hence, $\left(\text{B}\right)$ is the correct answer.