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3 years ago

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Light is incident perpendicularly on four transparent films of different thickness. The thickness of each film is given in the d

rawings in terms of the wavelength λfilm of the light within the film. The index of refraction of each film is 1.5, and each is surrounded by air. Which film(s) will appear bright due to constructive interference when viewed from the top surface, upon which the light is incident?

1 answer:

exis [7]3 years ago

6 0

Answer:

1/4 λ film

Explanation:

Let L = total path length then L = 2 t where t is film thickness

There will be a 180 deg phase change at the air-film interface but no

phase change at the film-air interface

L = n * wavelength / 2 where n = 1, 3, 5, 7 etc

(the total path L must be an odd number of 1/2 wavelengths)

Or t = n * wavelength / 4 (the film must be an odd number

of 1/4 wavelengths thick)

1/4 λ film satisfies this condition

Note: Find the missing diagram in the attachment section.

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The angle of separation is $\Delta \theta = 0.93 ^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta _ i = 56^o$

The refractive index of violet light in diamond is $n_v = 2.46$

The refractive index of red light in diamond is $n_r = 2.41$

The wavelength of violet light is $\lambda _v = 400nm = 400*10^{-9}m$

The wavelength of red light is $\lambda _r = 700nm = 700*10^{-9}m$

Snell's Law can be represented mathematically as

$\frac{sin \theta_i}{sin \theta_r} = n$

Where $\theta_r$ is the angle of refraction

=> $sin \theta_r = \frac{sin \theta_i}{n}$

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$sin \theta_r__{v}} = \frac{sin \theta_i}{n_v}$

substituting values

$sin \theta_r__{v}} = \frac{sin (56)}{2.46}$

$sin \theta_r__{v}} = 0.337$

$\theta_r__{v}} = sin ^{-1} (0.337)$

$\theta_r__{v}} = 19.69^o$

Now considering red light

$sin \theta_r__{R}} = \frac{sin \theta_i}{n_r}$

substituting values

$sin \theta_r__{R}} = \frac{sin (56)}{2.41}$

$sin \theta_r__{R}} = 0.344$

$\theta_r__{R}} = sin ^{-1} (0.344)$

$\theta_r__{R}} = 20.12^o$

The angle of separation between the red light and the violet light is mathematically evaluated as

$\Delta \theta = \theta_r__{R}} - \theta_r__{V}}$

substituting values

$\Delta \theta =20.12 - 19.69$

$\Delta \theta = 0.93 ^o$

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