How much faster, in meters per second, does light travel in sapphire than in diamond? See Table 33-1.

Short Answer

Expert verified

The difference in speed is 4.55×107ms.

Step by step solution

01

Definition of refractive index.

The refractive index of any medium signifies the bending ability of light in that perticular medium. Hence, the medium is more refractive, and the bending of the light in that medium will also be more.

02

Determine of change in speed of the light with the change of the medium.

From the table 33-1, we get the refractive index of diamond and sapphire as:

The refractive index of diamond isnd=2.42

The refractive index of sapphire isns=1.77

The formulae for calculating the change in velocity Δvwith the change in the medium are:

Δv=vs-vd=c1ns-1nd

Here, vsis the speed of light in the sapphire medium and vdthe speed of light in the diamond medium.

C is the speed of light in the vaccum.

Therefore,

Δv=3×10811.77-12.42=4.55×107ms

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Most popular questions from this chapter

In Fig. 35-4, assume that two waves of light in air, of wavelength 400nm, are initially in phase. One travels through a glass layer of index of refraction n1=1.60and thickness L. The other travels through an equally thick plastic layer of index of refraction n2=1.50. (a) What is the smallest value Lshould have if the waves are to end up with a phase difference of 5.65 rad? (b) If the waves arrive at some common point with the same amplitude, is their interference fully constructive, fully destructive, intermediate but closer to fully constructive, or intermediate but closer to fully destructive?

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 r3 and r4 interfere, 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,n2andn3, the type.

Of interference, the thin-layer thickness L in nanometres, and the wavelength λ in nanometres of the light as measured in air.

Where λ 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?

In Fig. 35-45, a broad beam of monochromatic light is directed perpendicularly through two glass plates that are held together at one end to create a wedge of air between them. An observer intercepting light reflected from the wedge of air, which acts as a thin film, sees 4001 dark fringes along the length of the wedge. When the air between the plates is evacuated, only 4000 dark fringes are seen. Calculate to six significant figures the index of refraction of air from these data.

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.

A double-slit arrangement produces interference fringes for sodium light (λ=589nm)that have an angular separation of 3.50×10-3rad. For what wavelength would the angular separation be 10% greater?

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