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A soap film appears red (of wavelength 650 nm) when viewed with an incident white light that is...

Question:

A soap film appears red (of wavelength 650 nm) when viewed with an incident white light that is perpendicular to the film. If the index of refraction of the film is 1.38, what is the minimum thickness of the film? Answer in units of nm.

Constructive Interference:

Generally, interference occur when waves such as light meet. Interference can be constructive or destructive depending on their phase difference. This phenomenon can occur when light passes through and interface such as in layers of thin films. In this case, the reflecting light rays at every interface may constructively interfere to produce waves of higher amplitude or destructively interfere to cancel each other.

Answer and Explanation:

Let the thickness of the soap film be t, and the index of refraction be n. To calculate for the minimum thickness, we use the following equation:

{eq}\displaystyle t = \frac{ \left(m+ \frac{1}{2} \right) \lambda}{2n} {/eq}

where m is the order of reflection. To solve for the minimum thickness, we use m=0. So we have,

{eq}\displaystyle t = \frac{ \left(0+ \frac{1}{2} \right) 650\ \rm nm}{2(1.38)} = 118\ \rm nm {/eq}

Therefore, the minimum thickness of the soap film to appear red is 118 nm.


Learn more about this topic:

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Constructive and Destructive Interference

from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 8 / Lesson 16
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