Question

1. A mirror of focal length 22.4 cm creates an image with magnification -0.350. What is the image distance?

2. As light travels between media with different indices of refraction, which of the following wave properties changes: speed, frequency, wavelength?

Answer 1

1). A mirror of focal length 22.4 cm creates an image with magnification -0.350. What is the image distance

f= 22.4 cm

M= -0.350

The mirror equation expresses the quantitative relationship between the object distance (d_o), the image distance (d_i), and the focal length (f). The equation is stated as follows:

      Equation .......................(1)

The magnification equation relates the ratio of the image distance and object distance to the ratio of the image height (h_i) and object height (h_o). The magnification equation is stated as follows:

So that by the given data

M= -d_i/d₀ = -0.350

             d_i = 0.350d₀       Equation .......................(2)

From the   Eq.............(1)

            d_i d₀ = (d_i + d₀ )f

From Eq .................(2)

             0.350*d₀² = d₀ (1+0.350)22.4

                => d₀ = (1.350*22.4)/0.350

                             = 86.4 cm.

by substituting the d₀ value in equation .............(2)

                       d_i = 0.350* 86.4

                            = 30.24 cm.

The image distance is 30.24 cm.

2). As light travels between media with different indices of refraction, which of the following wave properties changes: speed, frequency, wavelength?

Symbols used: ? is wavelength, ? is frequency, c,v are speeds of light in vacuum and in the medium.

Alright. First, we can look at just frequency and determine if frequency should change on passing through a medium.

Frequency can't change

Now, let's take a glass-air interface and pass light through it. (In SI units) In one second, ? "crest"s will pass through the interface. Now, a crest cannot be distroyed except via interference, so that many crests must exit. Remember, a crest is a zone of maximum amplitude. Since amplitude is related to energy, when there is max amplitude going in, there is max amplitude going out, though the two maxima need not have the same value.

Also, we can directly say that, to conserve energy (which is dependent solely on frequency), the frequency must remain constant.

Speed can change

There doesn't seem to be any reason for the speed to change, as long as the energy associated with unit length of the wave decreases. It's like having a wide pipe with water flowing through it. The speed is slow, but there is a lot of mass being carried through the pipe. If we constrict the pipe, we get a jet of fast water. Here, there is less mass per unit length, but the speed is higher, so the net rate of transfer of mass is the same.

In this case, since ??=v, and ? is constant, change of speed requires change of wavelength. This is analogous to the pipe, where increase of speed required decrease of cross-section (alternatively mass per unit length)

Why does it have to change?

Alright. Now we have established that speed can change, lets look at why. Now, an EM wave(like light), carries alternating electric and magnetic fields with it. Here's an animation. Now, in any medium, the electric and magnetic fields are altered due to interaction with the medium. Basically, the permittivities/permeabilities change. This means that the light wave is altered in some manner. Since we can't alter frequency, the only thing left is speed/wavelength (and amplitude, but that's not it as we shall see)

Using the relation between light and permittivity/permeability (?0?0=1/c2 and ??=1/v2), and ?=?r?0,?=?r?0,n=c/v (n is refractive index), we get n=?r?r?????, which explicitly states the relationship between electromagnetic properties of a material and its RI.

Basically, the relation ??=1/v2 guarantees that the speed of light must change as it passes through a medium, and we get the change in wavelength as a consequence of this.

Answers is speed and wave length will change.