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Chapter 5: Q32E (page 1182)

What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with wavelength 480 nm? The index of refraction of the film is 1.33, and there is air on both sides of the film.

Short Answer

Expert verified

The thickness of soap film is 180nm.

Step by step solution

01

Snell’s Law

We are given 位iar=480nm&n=1.33. The first reflected ray experiences a phase change. The red circle indicates the phase change.

But the second reflected ray, as you see below, experiences no phase change at all.

Hence, we have two reflected rays with one phase change and we need them to interact destructively when the soap illuminated by a 480 nm-light beam.

So, the thickness of the soap bubble is given by

2t=m位soapt=m位soap2 (1)

02

About Snell’s Law

We know, from Snell鈥檚 Law

n1位1=n2位2nair位air=nsoap位soap

Solving for位soap and thatnair=1

位soap=位airnsoap

Substitute into above equation

t=m位air2nsoap

Since the question asked to exclude the case the case of zero thickness, we will not use m = 1.0.

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