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
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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