/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q39P Light of wavelength 624 nm is in... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Light of wavelength 624 nm is incident perpendicularly on a soap film (n = 1.33) suspended in air. What are the (a) least and (b) second least thicknesses of the film for which the reflections from the film undergo fully constructive interference?

Short Answer

Expert verified

The least thickness is 117 nm, and the 2nd least thickness is 352 nm.

Step by step solution

01

Interference in thin films

Bright colors reflected from thin oil on water and soap bubbles are a consequence of light interference. Due to the constructive interference of light reflected from the front and back surfaces of the thin film, these bright colors can be seen. For a perpendicular incident beam, the maximum intensity of light from the thin film satisfies the condition:

2L=(m+12)λn2 â¶Ä‰â¶Ä‰m=0, 1, 2,...(Maxima—bright film in the air)

where λ is the wavelength of the light in air, L is its thickness, and n2 is the film’s refraction index.

02

Thickness of the soap film

The least and the 2nd least thickness are associated with m=0, 1the order of maximum intensity. The thickness corresponding to this order of maxima is

For m=0;

Lo=m+12λ2n2=0+12624 nm2(1.33)=117 nm

For m=1;

L1=m+12λ2n2=1+12624 nm2(1.33)=352 nm

Thus, the least thickness is 117 nm, and the 2nd least thickness is 352 nm.

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 91Ó°ÊÓ!

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

Add the quantities y1=10sinӬt, y2=15sin(Ӭt+30°)andy3=5sin(Ӭt-45°) using the phasor method

Transmission through thin layers. In Fig. 35-43, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) Part of the light ends up in material 3 as ray r3(the light does not reflect inside material 2) and r4(the light reflects twice inside material 2). The waves of and interfere, r3and r4here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35-3 refers to the indexes of refraction n1,n2and n3, the type of interference, the thin-layer thickness Lin nanometers, and the wavelength in nanometers of the light as measured in air. Where λis missing, give the wavelength that is in the visible range. Where Lis missing, give the second least thickness or the third least thickness as indicated.

The rhinestones in costume jewellery are glass with index of refraction 1.50. To make them more reflective, they are often coated with a layer of silicon monoxide of index of refraction 2.00.What is the minimum coating thickness needed to ensure that light of wavelength 560nm and of perpendicular incidence will be reflected from the two surfaces of the coating with fully constructive interference?

Figure 35-22 shows two light rays that are initially exactly in phase and that reflect from several glass surfaces. Neglect the slight slant in the path of the light inthe second arrangement.

(a) What is the path length difference of the rays?

In wavelengthsλ,

(b) what should that path length difference equal if the rays are to be exactly out of phase when they emerge, and

(c) what is the smallest value of that will allow that final phase difference?

A thin flake of mica (n = 1.58) is used to cover one slit of a double-slit interference arrangement. The central point on the viewing screen is now occupied by what had been the seventh bright side fringe (m = 7). If λ=550nm , what is the thickness of the mica?

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.