Physics, 05.05.2020 06:15 jacob12049

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 .

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Answer from: ummsumaiyah3583

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.

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 from: BigI80531

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 $10^{-5}$ ° $C^{-1}$

Answer from: Quest

5.5 n====== $\sum f = ma$ assuming that $\sum f = f_{app}$ we have m = 0.145 kg a = 38 m/s²therefore $f_{app} = (0.145\text{ kg})(38\text{ m/s}^2) = 5.51\text{ n}$ 5.51 n, but two significant figures.final answer: 5.5 n

Answer from: Quest

Answer (d) the universe had cooled enough for atoms to combine, protons and neutrons were too far apart to form nuclei;

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A thin uniform film of refractive index 1.750 is placed on a sheet of glass with a refractive index...