Solutions to Assignment 10

FGT 35.30

The power of a laser is P=0.4mW. This is the rms power. The intensity is related to the power by, I = P/Area. The intensity is related to the peak electric field by $I = \epsilon_0cE_0^2/2$, and the peak magnetic field is related to the peak electric field by, B₀ = E₀/c. When the laser is focused on a spot of diameter d = 1.2mm, we then find, E₀ = 516N/C, $B_0 = 1.7\mu T$. However when the laser is focused on a spot of diameter d=633nm, the peak electric and magnetic fields are, $E_0 = 9.8\times 10^5N/C$, B₀ = 3.25mT.

FGT 35.39

The momentum transfer when a particle of momentum p_i reflects elastically at angle $\theta$ (to the normal) from a surface, is given by, $\delta p_i = 2 p_i cos(\theta)$. The momentum flux (= radiation pressure) is then, $2 u cos(\theta)$. We are asked for the momentum transfer, so we must multiply by the area over which the pressure acts and the time over which it acts, so that, momentum transfer $= 2 u cos(\theta) A \delta t$.

FGT 35.52

A polarizer consists of conducting wires. Light of polarization parallel to the direction of the conducting wires is absorbed. Light perpendicular to the conducting wires is transmitted. The polarization axis of the transmitted light is then perpendicular to the direction of the wires. The axis of a polarizer is then defined to be perpendicular to the conducting wires so that light which is polarized parallel to the polarizer axis is transmitted. With this definition, we find that,

I₂ = I₀ cos²(47)cos²(63) = 0.096 (1)

FGT 35.64

In the summary to the chapter the formula $tan(\theta_B) = n$ is given. This formula applies to the case of light incident through air and reflecting off a material with refractive index n. The general formula is,

$$tan(\theta_B) = n_2/n_1$$ (2)

for light incident from medium ``1''. In the case of a light source underwater, medium ``1'' is water and medium ``2'' is air. We then have, $tan(\theta_B) = 1/1.33$from which we find, $\theta_B=37^o$.

FGT 35.65

The amount of energy absorbed is given by,

E = 800*0.4*0.5*3600 J = 0.576 MJ (3)

The latent heat of vaporisation is L=41kJ/mole, so the number of moles of water evaporated is,

n = E/L = 14.1 moles (4)

The molar mass of water is 18.01g/mole, so the mass of water evaporated is m = 14.1*18.01 gm = 0.253kg. (if you are on the beach all day it is easy to see how you get dehydrated).

FGT 36.26

Because the screen is so far from the prism, the majority of the distance between the blue and red spots is determined by the angles the two colors make when exiting the prism. We thus find these angles.

We find the outgoing angle $\alpha$ from using Snell's law at the first surface,

$$\theta_1 = 90-65, \ \ \ \theta_2 = sin^{-1}[{n_1\over n_2}sin(\theta_1)]$$ (5)

and Snell's law at the second surface,

$$\theta_3 = 50-\theta_2\ \ \ \theta_4 = sin^{-1}[{n_2\over n_1}sin(\theta_3)]$$ (6)

and finally $\alpha = \theta_4-25$

Plugging the numbers for blue light, with n_b=1.528 yields, $\theta_2 = 16.06^o$, $\theta_4 = 58.55^o$, $\alpha_b = 33.55^o$, while for red light, with n_r = 1.514 we find, $\theta_2 = 16.21^o$, $\theta_4 = 57.35^o$, $\alpha_b = 32.35^o$.

The location of the beams on a screen which is distance D=10m from the prism is found from $d = Dtan(\alpha)$. Plugging the numbers for blue light gives d_b=6.631m, while for red light we find, d_r = 6.334m. The distance between the two spots on the screen is then $\delta d = d_b=d_r = 0.30m$.

FGT 36.27

In this problem we have to take account of the displacement of the beam inside the prism as well as outside, because the screen is close to the prism. Consider a co-ordinate system centered at the apex of the prism, with the x-axis bisecting the prism and the y-axis upward. The x distance a beam travels inside the prism is given by, x(y) = 0.866(1-y/1.5)where all distances are in cm. We use Snell's law at the first surface, where $\theta_1 = 30^o$, and so,

$$\theta_2 = sin^{-1}[{1\over 1.5}sin(30)] = 19.47^o$$ (7)

The angle of incidence (to the normal) at the second surface, $\theta_3 = 30-\theta_2 = 10.53^o$. Using Snell's law at the second surface yields,

$$\theta_4=sin^{-1}[1.5 sin(10.53)] = 15.91^o$$ (8)

Now consider two rays. Firstly a ray incident at the very top of the prism. This ray still refracts at both interfaces, but the distance it travels inside the prism is negligible. This spot this ray makes on the screen is then displaced by,

d₁ = 2cm*tan(15.91)= 0.57cm (9)

from the top of the prism. Secondly consider a ray incident just above the center of the prism. This ray is displaced inside prism and after it leaves the prism. The total displacement is,

d₂ = 0.866*tan(10.53) + 2cm*tan(15.91) = 0.73cm (10)

The same analysis applies to the same two rays incident on the bottom half of the prism. The illumination pattern on the screen is then seen to be: There are dark regions of size 0.57cm at the top and bottom edges of the prism. There is a central region of size 1.46 which is illuminated by both halves of the prism. The remaining sections are ``single'' illuminated.

FGT 36.51

Rays which undergo total internal reflection at the water/air interface cannot be seen by an observer above the water. The condition for total internal reflection at the water/air interface is,

$$\theta_c = sin^{-1}[{1\over n_w}] = 48.75^o$$ (11)

The cork has radius 1.5cm and it has a section of depth 1.2cm below the surface of the water. The rod which is depth l below the surface, produces a light ray which is at angle of incidence $\theta_1$(to the normal) to the water/air surface. If $\theta_1>\theta_c$, total internal reflection occurs. The longest rod which is obscured thus leads to a ray at angle $\theta_c$ from its tip. The geometry of the situation implies that,

$$tan(\theta_c) = {1.5+x\over 1.2+l}$$ (12)

where all lengths are in cm. x is related to the critical angle by, $x=1.2tan(\theta_c) = 1.368cm$. Solving for l yields,

$$l = {2.868 \over tan(48.75)}-1.2 = 1.315cm$$ (13)

Notice that you also get this result if you just write

$$tan(\theta_c) = 1.5/l$$ (14)

However you need to give a discussion as to why this also works.

FGT 36.52

The incident and outgoing rays are at angle 30^o to the normal to the surfaces of the slab. However inside the slap the ray is at angle $\theta_2$ to the normal. $\theta_2$ is found from Snell's law,

$$sin(30) = 1.52sin(\theta_2)$$ (15)

so that, $\theta_2=19.2^o$. The vertical displacement between the point of entry of the ray and the point of exit of the ray is, $y_1 = 2 tan(\theta_2) = 0.696cm$, where t=2cm is the thickness of the slab. If the ray had not been refracted, the exit point would have been displaced vertically by, y₂ = 2.0tan(30) = 1.155cm. The difference in vertical displacement is then y = y₂-y₁ = 0.459cm. The perpendicular distance between the two outgoing rays is then, d=0.459cos(30) = 0.397cm.