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Chapter 35: Q96P (page 1080)

A camera lens with index of refraction greater than 1.30 is coated with a thin transparent film of index of refraction 1.25 to eliminate by interference the reflection of light at wavelength λ that is incident perpendicularly on the lens. What multiple of λgives the minimum film thickness needed?

Short Answer

Expert verified

Thus, the minimum film needed is 0.200.

Step by step solution

01

The thin film interference of a coating on glass lens.

The formula for the thin film interference of coating on the glass lens:

Lmin=λ4n2

02

The minimum film thickness needed by multiple of .

Use the above formula as follows:

Lmin=λ4n2=λ41.25=0.200λ

Further solve as follows:

Lmin=0.200λLminλ=0.200

Hence, the minimum film needed is 0.200.

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Most popular questions from this chapter

Figure 35-28 shows four situations in which light reflects perpendicularly from a thin film of thickness L sandwiched between much thicker materials. The indexes of refraction are given. In which situations does Eq. 35-36 correspond to the reflections yielding maxima (that is, a bright film).

Reflection by thin layers. In Fig. 35-42, 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.) The waves of rays r1and r2interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35- 2 refers to the indexes of refraction n1, n2andn3, the type of interference, the thin-layer thickness Lin nanometres, and the wavelength λin nanometres 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.

Monochromatic green light, of wavelength 500 nm, illuminates two parallel narrow slits 7.70 mm apart. Calculate the angular deviation ( θin Fig. 35-10) of the third-order (m=3)bright fringe (a) in radians and (b) in degrees.

A thin film of liquid is held in a horizontal circular ring, with air on both sides of the film. A beam of light at wavelength 550 nm is directed perpendicularly onto the film, and the intensity I of its reflection is monitored. Figure 35-47 gives intensity I as a function of time the horizontal scale is set by ts=20.0s. The intensity changes because of evaporation from the two sides of the film. Assume that the film is flat and has parallel sides, a radius of 1.80cm, and an index of refraction of 1.40. Also assume that the film’s volume decreases at a constant rate. Find that rate.

In Figure 35-50, two isotropic point sources S1and S2emit light in phase at wavelength λand at the same amplitude. The sources are separated by distance d=6.00λon an x axis. A viewing screen is at distance D=20.0λfrom S2and parallel to the y axis. The figure shows two rays reaching point P on the screen, at height yp. (a) At what value of do the rays have the minimum possible phase difference? (b) What multiple of λgives that minimum phase difference? (c) At what value of ypdo the rays have the maximum possible phase difference? What multiple of λgives (d) that maximum phase difference and (e) the phase difference when yp=d? (f) When yp=d, is the resulting intensity at point P maximum, minimum, intermediate but closer to maximum, or intermediate but closer to minimum?
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