Math  /  Trigonometry

QuestionA soap film (index of refraction 1.33) is surrounded on both sides by air. When white light shines nearly perpendicularly on the film, you see bright interference colours of wavelength =478 nm=478 \mathrm{~nm} in the reflected light. What is the second smallest possible value for the thickness of the film?

Studdy Solution

STEP 1

1. The soap film is surrounded by air on both sides.
2. The index of refraction of the soap film is n=1.33 n = 1.33 .
3. The observed wavelength of light in air is λ=478nm \lambda = 478 \, \text{nm} .
4. We are looking for the second smallest possible thickness of the film that results in constructive interference.

STEP 2

1. Understand the condition for constructive interference in thin films.
2. Calculate the wavelength of light within the film.
3. Determine the formula for the thickness of the film.
4. Calculate the second smallest thickness.

STEP 3

Understand the condition for constructive interference in thin films:
For constructive interference in a film surrounded by air, the condition is:
2nt=mλ 2nt = m\lambda
where n n is the index of refraction, t t is the thickness of the film, m m is the order of interference (an integer), and λ \lambda is the wavelength of light in air.

STEP 4

Calculate the wavelength of light within the film:
The wavelength of light in the film, λ \lambda' , is given by:
λ=λn \lambda' = \frac{\lambda}{n} λ=478nm1.33 \lambda' = \frac{478 \, \text{nm}}{1.33}

STEP 5

Determine the formula for the thickness of the film:
Rearrange the constructive interference condition to solve for t t :
t=mλ2n t = \frac{m\lambda}{2n}

STEP 6

Calculate the second smallest thickness:
For the smallest thickness, m=1 m = 1 . For the second smallest thickness, m=2 m = 2 .
t=2×478nm2×1.33 t = \frac{2 \times 478 \, \text{nm}}{2 \times 1.33}
Calculate:
t=956nm2.66 t = \frac{956 \, \text{nm}}{2.66}
t359.4nm t \approx 359.4 \, \text{nm}
The second smallest possible value for the thickness of the film is:
359.4nm \boxed{359.4 \, \text{nm}}

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