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?

Short Answer

Expert verified

The speed of light in material 1 is 2×108m/s.

Step by step solution

01

Given information

  1. The incident angle of a beam of light in material 1 is,θ1=30° .
  2. The angle of refraction of light in material 2 is,θ2 .
  3. The index of refraction of material 1 is,n1.
  4. The index of refraction of material 2 is,n2 .
02

Diffraction of light

If a ray of light passes through multiple layers of mediums, then it deviates from its direct path, this deviation of light due to mediums is described as ‘diffraction of light’.

The law used to establish a relationship between the angles of incident and refracted waves of light is the Snell’s law.

03

Speed of light in material 1

According to the Snell's law (the law of refraction),

n1sinθ1=n2sinθ2

If there is only material 1, then,θ1=θ2the expression reduces to,

n1=n2

From the graph at θ2=30°, the value ofn2is,n2=1.5.

So, the value of n1will be,n1=1.5

The formula for the speed with which the light propagates in medium 1 is given by,

v=cn1

Here, speed of lightc=3×108m/s.

Putting values,

v=3×108m/s1.5v=2×108m/s

Hence, the speed of light in material 1 is2×108m/s .

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Most popular questions from this chapter

The lens in a Newton’s rings experiment (see problem 75) has diameter 20 mm and radius of curvature R=5.0m. For A=589nm in air, how many bright rings are produced with the setup (a) in air and
(b) immersed in water (n=1.33)?

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.

In Fig. 35-39, two isotropic point sources S1 and S2 emit light in phase at wavelength λ and at the same amplitude. The sources are separated by distance 2d=6λ. They lie on an axis that is parallel to an x axis, which runs along a viewing screen at distance D=20.0λ. The origin lies on the perpendicular bisector between the sources. The figure shows two rays reaching point P on the screen, at positionxP. (a) At what value of xPdo the rays have the minimum possible phase difference? (b) What multiple ofλ gives that minimum phase difference? (c) At what value ofxPdo the rays have the maximum possible phase difference? What multiple of λ gives (d) that maximum phase difference and (e) the phase difference when xP=6λ ? (f) When xP=6λ, is the resulting intensity at point P maximum, minimum, intermediate but closer to maximum, or intermediate but closer to minimum?

Suppose that Young’s experiment is performed with blue-green light of wavelength 500 nm. The slits are 1.20 mm apart, and the viewing screen is 5.40 m from the slits. How far apart are the bright fringes near the center of the interference pattern?

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.

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