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A 600 nm thick soap film (n=1.40) in air is illuminated with white light in a direction...

Question:

A 600 nm thick soap film (n=1.40) in air is illuminated with white light in a direction perpendicular to the film. For how many different wavelengths in a 300 to 700 nm range, is there

a) fully constructive interference and

b) fully destructive interference in the reflected light?

Interference:

When two waves of similar type get superimposed with each other to form a new wave is called interference. The new wave formed can be of different amplitude or same amplitude. There are two types of interference.

Answer and Explanation:

Given data


  • The redfractive index of soap film is, {eq}n = 1.4 {/eq}.
  • The range of wavelength is, {eq}300\;{\rm{nm}} {/eq} to {eq}700\;{\rm{nm}} {/eq}.
  • The thickness of film is, {eq}L = 600\;{\rm{nm}} {/eq}.


The expression for constructive interference in a oil film is given as,

{eq}2L = \left( {m + \dfrac{1}{2}} \right)\dfrac{\lambda }{n} {/eq}

Rearranging the above expression to determine the wavelength,

{eq}\lambda = \dfrac{{4Ln}}{{2m + 1}} {/eq}


Here, values of m will be 1, 2, 3 ....


Substituting the values, in above expression.

For m = 0,

{eq}\begin{align*} \lambda &= \dfrac{{4\left( {600} \right) \times 1.4}}{{2\left( 0 \right) + 1}}\\ \lambda &= 3360\;{\rm{nm}} \end{align*} {/eq}


For m = 1,

{eq}\begin{align*} \lambda &= \dfrac{{4\left( {600} \right) \times 1.4}}{{2\left( 1 \right) + 1}}\\ \lambda &= 1120\;{\rm{nm}} \end{align*} {/eq}


For m = 2,

{eq}\begin{align*} \lambda &= \dfrac{{4\left( {600} \right) \times 1.4}}{{2\left( 2 \right) + 1}}\\ \lambda &= 672\;{\rm{nm}} \end{align*} {/eq}


For m = 3,

{eq}\begin{align*} \lambda &= \dfrac{{4\left( {600} \right) \times 1.4}}{{2\left( 3 \right) + 1}}\\ \lambda &= 480\;{\rm{nm}} \end{align*} {/eq}


For m = 4,

{eq}\begin{align*} \lambda &= \dfrac{{4\left( {600} \right) \times 1.4}}{{2\left( 4 \right) + 1}}\\ \lambda &= 373.33\;{\rm{nm}} \end{align*} {/eq}


For m = 5,

{eq}\begin{align*} \lambda &= \dfrac{{4\left( {600} \right) \times 1.4}}{{2\left( 5 \right) + 1}}\\ \lambda &= 305.45\;{\rm{nm}} \end{align*} {/eq}


Thus, there are four values of wavelength between {eq}300\;{\rm{nm}} {/eq} to {eq}700\;{\rm{nm}} {/eq}.


(b)

The expression for condition of full destructive interference in oil film is given as,

{eq}2L = \dfrac{{m\lambda }}{n} {/eq}

Rearranging for the values of wavelength,

{eq}\lambda = \dfrac{{2Ln}}{m} {/eq}


Substituting the values of m as 1, 2, 3 ...

For m = 1,

{eq}\begin{align*} \lambda &= \dfrac{{2\left( {600} \right) \times 1.4}}{1}\\ \lambda &= 1680\;{\rm{nm}} \end{align*} {/eq}


For m = 2,

{eq}\begin{align*} \lambda &= \dfrac{{2\left( {600} \right) \times 1.4}}{2}\\ \lambda &= 840\;{\rm{nm}} \end{align*} {/eq}


For m = 3,

{eq}\begin{align*} \lambda &= \dfrac{{2\left( {600} \right) \times 1.4}}{3}\\ \lambda &= 560\;{\rm{nm}} \end{align*} {/eq}


For m = 4,

{eq}\begin{align*} \lambda &= \dfrac{{2\left( {600} \right) \times 1.4}}{4}\\ \lambda &= 420\;{\rm{nm}} \end{align*} {/eq}


For m = 5,

{eq}\begin{align*} \lambda &= \dfrac{{2\left( {600} \right) \times 1.4}}{5}\\ \lambda &= 336\;{\rm{nm}} \end{align*} {/eq}


So, there are three values of wavelength between {eq}300\;{\rm{nm}} {/eq} to {eq}700\;{\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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