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Chapter 35: Q100P (page 1080)

A thin film suspended in air is 0.410 渭尘thick and is illuminated with white light incident perpendicularly on its surface. The index of refraction of the film is 1.50. At what wavelength will visible light that is reflected from the two surfaces of the film undergo fully constructive interference?

Short Answer

Expert verified

Thus, the wavelength of visible light is 492nm.

Step by step solution

01

According to the question.

The formula for the constructive interference:

2n2L=(m+12)位

Here,位is wavelength of light in air/vacuum.

02

The wavelength of visible light. 

Use the above formula as follows:

位=2n2Lm+12=21.50410鈥�nmm+12=1230nmm+12

Where m=0,1,2,...The only value of mwhich, when substituted into equation above, would yield a wavelength that within the visible light range ism=1 .

Therefore, the wavelength is:

位=1230鈥�nm1+12=492鈥�nm

Hence, the wavelength of visible light is492nm.

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Most popular questions from this chapter

Two waves of the same frequency have amplitudes 1.00 and 2.00. They interfere at a point where their phase difference is 60.0掳. What is the resultant amplitude?

Suppose that the two waves in Fig. 35-4 have a wavelength 位=500nmin air. What multiple of 位gives their phase difference when they emerge if (a) n1=1.50, n2=16and L=8.50渭m; (b) n1=1.62, n2=1.72, and L=8.50渭m; and (c) n1=1.59, n2=1.79, and L=3.25渭m? (d) Suppose that in each of these three situations, the waves arrive at a common point (with the same amplitude) after emerging. Rank the situations according to the brightness the waves produce at the common point.

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 位in nanometres of the light as measured in air. Where 位is 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 ( 胃in Fig. 35-10) of the third-order (m=3)bright fringe (a) in radians and (b) in degrees.

Find the slit separation of a double-slit arrangement that will produce interference fringes0.018鈥�radapart on a distant screen when the light has wavelength位=589鈥�nm.
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