Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What is the thinnest film of a coating with \(n = 1.42\) on glass \((n = 1.52)\) for which destructive interference of the red component \((650nm)\) of an incident white light beam in air can take place by reflection?

Short Answer

Expert verified

The thinnest film of a coating is 114nm.

Step by step solution

01

Formulas used to solve the question

Destructive interference

2t=(m+12)λ

Snell’s law:

n1λ1=n2λ2

02

Determine the thickness in terms of λ

Given: nair=1.0

ncoating=1.42nglass=1.52λair=650nm=650*10-9m

Light beam reflects from a surface that has a greater index of refraction than the index of refraction of the first medium, the wave experiences a phase change. This means that the first ray experiences the phase change and the second ray also experiences a phase change.

From equation (1), at m = 0 for the thinnest coating

t=14λ (3)

03

Determine the thickness

From equation (2),

\({n_{coating}}{\lambda _{coating}} = {n_{air}}{\lambda _{air}}\)

\( \Rightarrow {\lambda _{coating}} = \frac{{{n_{air}}{\lambda _{air}}}}{{{n_{coating}}}}\)

Plug the given,

\( \Rightarrow {\lambda _{coating}} = \frac{{{\lambda _{air}}}}{{{n_{coating}}}}\)

Plug into (3),

\(t = \frac{{{\lambda _{air}}}}{{4{n_{coating}}}} = \frac{{650*10{}^{ - 9}}}{{4*1.42}} = 114nm\)

Thus, the thinnest film of a coating is \(114nm\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The explanation given in Section 33.6 for the color of the setting sun should apply equally well to the risingsun, since sunlight travels the same distance through the atmosphere to reach your eyes at either sunrise or sunset. Typically, however, sunsets are redder than sunrises. Why? (Hint:Particles of all kinds in the atmosphere contribute to scattering.)

It has been proposed that automobile windshields and headlights should have polarizing filters to reduce the glare of oncoming lights during night driving. Would this work? How should the polarizing axes be arranged? What advantages would this scheme have? What disadvantages?

34.15 The thin glass shell shown in Fig. E34.15 has a spherical shape with a radius of curvature of 12cm, and both of its surfaces can act as mirrors. A seed high is placed 15.0cmfrom the center of the mirror along the optic axis, as shown in the figure. (a) Calculate the location and height of the image of this seed. (b) Suppose now that the shell is reversed. Find the location and height of the seed’s image.

The two sourcesS1andS2shown in Fig. 35.3 emit waves of the same wavelength λand are in phase with each other. Suppose S1is a weaker source, so that the waves emitted by S1have half the amplitude of the waves emitted by S2. How would this affect the positions of the antipodallines and nodal lines? Would there be total reinforcement at points on the antipodal curves? Would there be total cancellation at points on the nodal curves? Explain your answers.

A spherical air bubble in water can function as a lens. Is it a converging or diverging lens? How is its focal length related to its radius?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free