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Chapter 35: Q9Q (page 1073)

Light travels along the length of a 1500 nm-long nanostructure. When a peak of the wave is at one end of the nanostructure, is there a peak or a valley at the other end of the wavelength (a) 500nm and (b) 1000nm?

Short Answer

Expert verified

(a)There is a ‘peak’at the other end of the wave.

(b) There is a ‘valley’ at the other end of the wave

Step by step solution

01

Given information

The total length of the nanostructure is, 1500nm.

The wavelength for case (a) is, 500nm.

The wavelength for case (b) is, 1000nm.

02

Wavelength

For a longitudinal wave, its wavelength can be defined as the distance between its two consecutive peaks or valleys.

The wavelength of any wave relies upon factors like wave frequency, wave propagation speed, and the propagating medium.

03

Step 3(a): The other end of the wave 

According to the question, the peak of the wave is at one end of the nanostructure, and also the wavelength is the distance between two peaks and valleys of a wave.

Then, the wave showing peak and valley can be drawn as,

Hence, there is a ‘peak’ at the other end of the wave

04

Step 4(b): The other end of the wave 

Similarly, for a wavelength value of 1000nm, the wave having a peak at one end can be drawn as,

Hence, there is a ‘valley’ at the other end of the wave.

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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 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 n3the 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.

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 r3and r4interfere, 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 n3, the type of interference, the thin-layer thickness L in 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.

A double-slit arrangement produces interference fringes for sodium light(λ=589nm)that are 0.200Capart. What is the angular separation if the arrangement is immersed in water (n=1.33)?

In Fig. 35-32a, a beam of light in material 1 is incident on a boundary at an angle of 30o. The extent to which the light is bent due to refraction depends, in part, on the index of refraction n2of material 2. Figure 35-32b gives the angle of refraction θ2versus n2for a range of possible n2values, from na=1.30to nb=1.90. What is the speed of light in material 1?

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°. What is the resultant amplitude?
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