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 ( 18.8 ∘C), this film is just thick enough for light with a wavelength 580.9 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 588.2 nm .

Answer 1

Complete 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 ( 18.8 ∘C), this film is just thick enough for light with a wavelength 580.9 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 588.2 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.

Answer:

the coefficient of linear expansion of the film is $\alpha = 7.93 *10^{-5} / ^oC$

Explanation:

  From the question we are told that

     The refractive index of the film is  $n_f = 1.750$

      The refractive index of the glass is $n_g = 1. 50$

       The wavelength of light reflected at 18°C is $\lambda _r = 580.9nm = 580.9*10^{-9}m$

      The wavelength of light reflected at 170°C is $\lambda_h = 588.2 nm = 588.2 * 10^{-9}m$

For destructive interference the condition is  

         $2t = \frac{m \lambda }{n_f}$

  Where m is the order of interference

              t is the thickness

               

For the smallest thickness is  when m= 1 and this is represented as

             $t = \frac{\lambda }{2n_f }$

At 18°C  the thickness would be

              $t_{r} = \frac{580.9 *10^{-9}}{2 * 1.750}$

              $t_{r} = 166nm$\

At 170° the  thickness is  

               $t_h = \frac{588.2 *10^{-9}}{2 * 1.750}$

               $t_h = 168 nm$

The coefficient of linear expansion f the film is mathematically represented as

              $\alpha = \frac{t_h - t_r}{t_r \Delta T}$

Substituting value

                 $\alpha = \frac{168 *10^{-9} - 166 *10^{-9} }{166*10^{-9} * (170 -18)}$

                 $\alpha = 7.93 *10^{-5} / ^oC$

Answer 2

Answer:

α=8.37 x $10^{-5}$ °$C^{-1}$

Explanation:

Complete question : What is the coefficient of linear expansion of the film?

SOLUTION:

There is a net ( λ/2 ) phase change due to reflection for this film, therefore, destructive interference is given by

2t = m( λ/n)   where n=1.750

for smallest non-zero thickness

t= λ/2n

At 18.8°C, $t_{o$=580.9 x $10^{-9}$/(2 x 1.750)

$t_{o$= 165.9nm

At 170°C, t= 588.2x $10^{-9}$/(2x1.750)

t=168nm

t=$t_{o$(1 + αΔT)

=>α= (t-$t_{o$)/ ($t_{o$ΔT)   [ΔT= 170-18.8 =151.2°C]

α= (168 x $10^{-9}$ - 165.9 x $10^{-9}$)/ (165.9 x $10^{-9}$ x 151.2)

α= 2.1 x $10^{-9}$/ 2.508 x $10^{-5}$

α=8.37 x $10^{-5}$ °$C^{-1}$

Therefore, the coefficient of linear expansion of the film is 8.37 x $10^{-5}$ °$C^{-1}$