A thin layer of oil (with index of refraction 1.18) is floating on water. What is the minimum...

Question:

A thin layer of oil (with index of refraction 1.18) is floating on water. What is the minimum thickness of the oil in the region that reflects green light (of wavelength 522 nm)? Answer in units of nm.

Interference:

Light refracts when passes through an interface such as in thin film layers. When this happens, some parts of the light are transmitted to the next layer while some are reflected off the surface. The reflected rays interfere with each other depending on their phase difference.

Answer and Explanation:

In situations where light rays are reflected on the interfaces formed by layered films, constructive interference occur when the following condition is satisfied:

{eq}\displaystyle t = \frac{m \lambda }{2n} {/eq}

In this case, t is the minimum thickness which is obtained when the order is m=1. Furthermore, {eq}\lambda = 522\ \rm nm {/eq} is the wavelength of the reflected light and {eq}n = 1.18 {/eq} is the index of refraction of oil.


Solving the minimum thickness yields:

{eq}\displaystyle t = \frac{(1) (522\ \rm nm) }{2(1.18)} = 221\ \rm nm {/eq}


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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