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We wish to coat flat glass (n = 1.50) with a transparent material (n = 1.25) so that reflection of light at wavelength 600 nm is eliminated by interference. What minimum thickness can the coating have to do this?

Short Answer

Expert verified

The minimum thickness of coating is1200nm

Step by step solution

01

Identification of given data

The index of refraction for coat flat glass is nc=1.50

The index of refraction for transparent material is role="math" localid="1663133484448" nc=1.25

The wavelength of light isλ=600nm

02

Understanding the concept

The index of refraction is the ratio of the speed of light in medium to speed of light in vacuum.

03

Determination of minimum thickness of coating

The minimum thickness of coating is given as:

t=λ2nc-nm

Substitute all the values in equation.

t=600nm21.50-1.25t=1200nm

Therefore, the minimum thickness of coating is1200nm

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Most popular questions from this chapter

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 r3andr4interfere, and here 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 n3the type of interference, the thin-layer thickness Lin nanometers, and the wavelength λin 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.

White light is sent downward onto a horizontal thin film that is sandwiched between two materials. The indexes of refraction are 1.80for the top material, 1.70for the thin film, and 1.50for the bottom material. The film thickness is5×10-7m . Of the visible wavelengths (400 to 700nm ) that result in fully constructive interference at an observer above the film, which is the (a) longer and (b) shorter wavelength? The materials and film are then heated so that the film thickness increases. (c) Does the light resulting in fully constructive interference shift toward longer or shorter wavelengths?

In Fig. 35-4, assume that the two light waves, of wavelength 620nm in air, are initially out of phase by π rad. The indexes of refraction of the media are n1=1.45 andn2=1.65 . What are the (a) smallest and (b) second smallest value of Lthat will put the waves exactly in phase once they pass through the two media?

If you move from one bright fringe in a two-slit interference pattern to the next one farther out,

(a) does the path length difference Lincrease or decrease and

(b) by how much does it change, in wavelengths λ ?

A Newton’s rings apparatus is to be used to determine the radius of curvature of a lens . The radii of the nth and (n+20th)bright rings are found to be 0.162cm and 0.368cm, respectively, in light of wavelength 546nm. Calculate the radius of curvature of the lower surface of the lens.

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