Assignment 1, solutions:

Problem 1:

(a)      Power P = (3 mJ)/(1 ns) = 3 10⁶ W.
          Irradiance E = P/A = P/(pr²) = (3 10⁶ W)/(p(15 10^-6)²) = 4.24 10¹⁵ W/m².
(b)      Energy = E A Dt = (4.24 10¹⁵ W/m²)(p(3 10^-10)²)(10^-9 s) = 1.2 10^-12 J.

Problem 2:

10⁴ J/s * 10^-15 s = 10^-11 J = ~ energy in each pulse.
hf = hc/l = energy per photon.
# of photons ~ 10^-11 J * l/(hc) ~ 10⁷.

Problem 3:

nsinq_t = sinq_i  (law of refraction).
geometry:
cosq_t = t/y.
sin(q_i-q_t) = x/y.
q_t = sin^-1(sinq_i/n) = 19.47^o.
y = t/cosq_t =1.0607 cm.
x = y sin(q_i-q_t) = y sin(10.53^o) = 0.00194 m.
x = 1.94 mm.

Problem 4:

The critical angle for TIR at the water-air interface is
q_t = sin^-1(n_air/n_water) = sin^-1(1/1.33) = 48.75^o.
Let d be the separation between the fish and the eye. 
Then tan(48.75^o) = (d/2)/1, d = 2.28 m.

Problem 5:

q_i = 30^o, nsinq_t = sinq_i, q_t = 18.21^o  (law of refraction).
Look at the triangle with angles q_t, 90^o + a_i, and 30^o. q_t + 90^o + a_i + 30^o = 180^o.
a_i = 180^o - 90^o - q_t - 30^o = 41.8^o.  (Sum of angles in triangle = 180^o.)
The critical angle is q_t =  sin^-1(1/1.6) = 38.7^o.  We have total internal reflection.
Now look at the triangle with angles b_i, 120^o and a_i.
b_i = 180^o - 120^o - a_i = 18.21^o.  (Sum of angles in triangle = 180^o.)
b_t = sin^-1(nsina_i) = 30^o  (law of refraction).
Symmetry:  b_t = q_i.
This symmetry is easily revealed when one rotates the diagram.  The vertical line mirrors all the relevant angles.

Problem 6:

Concave mirror: f is positive
1/x_o + 1/x_i = 1/f,  1/x_i = 1/20 - 1/5 = -3/20,  x_i = -20/3.
The image is located 6.66 cm behind the mirror.

Problem 7:

The image formed by the lens is upright.  It therefore is a virtual image.  The image is smaller than the object, it therefore is formed by a diverging lens.
M = -x_i/x_o = 0.5,  x_i = -0.5 x_o, image and object are on the same side of the lens.
x_o + x_i = 10 cm,  x_o - 0.5 x_o = 10 cm x_o = 20 cm
1/x_o + 1/x_i = 1/f,  1/20 - 1/10 = 1/f,  f = -20 cm

Tracing rays:

Draw the object and the image as arrows, with their tails on the optic axis. The image arrow has half the length of the object arrow.  They are separated by a distance d.  Draw the ray that connects the the tips of the arrows.  It intersects the optic axis a distance 2d from the object.  Draw the ray parallel to the optic axis.  It diverges after passing through the plane containing the lens.  The backward extension of this diverging ray must pass through the image.  Where this backward extension intersects the optic axis is the image focus of the lens.  The image focus is a distance 2d to the left of the lens, at the location of the object.

Problem 8: (E)

sinq = n(l)sinq',  sinq' = sinq/n(l), 
dsinq'/dq' = cosq',  dsinq'/dl =-[sinq/n(l)²]dn(l)/dl
Divide:
|dq'/dl| =[sinq/(cosq'n(l)²)]dn(l)/dl = [tanq'/n(l)]dn(l)/dl,
dq' = [tanq'/n(l)][dn(l)/dl]dl.

Problem 9: (A)

For the first lens: 1/40+ 1/x_i = 1/20, x_i = 40.  The image of the first lens becomes the object for the second lens.  The object distance is therefore -10 cm. 
For the second lens:  1/(-10) + 1/x_i = 1/10, x_i = 5.

Problem 10: (E)

For the concave mirror:  1/x_o + 1/x_i = 1/f,  f = R/2 = positive.  If x_o < f then x_i is negative, we have a virtual image.

Problem 11: (B)

E = hf - f = hc/500nm - 2.28 eV = 1242 eV-nm/500 nm - 2.28 eV = 0.2 eV.

Problem 12: (A)

Problem 13: (D)

Problem 14: (D)