Question 1

1A) A solid cube with sides of length 2a and refractive index nhas a point source of light at its center. Light is emitted from the point source uniformly into the cube, and the cube itself is floating in free space.

i)[7 pts] Find an expression for the largest value of the refractive index n for which none of the light emitted from the point source undergoes total internal reflection. Assume 100 percent transmission of all light that hits the surface of the cube UNLESS it is totally internally reflected.

The distance between the center of the cube and any corner of the cube is

$$d~=~\sqrt{a^2+a^2+a^2}~=~\sqrt{3}a\ \ \ .$$

The distance from the center of a face to a nearest corner of the cube is

$$h~=~\sqrt{a^2+a^2}~=~\sqrt{2} a\ \ \ .$$

Thus, the angle between a line drawn from the center of the cube to an edge and a line normal to the face is

$$\sin\theta~=~\sqrt{2\over 3}\ \ \ .$$

and hence the largest refractive index for which all light escapes (given a 100 percent transmission coefficient) is

$$n~=~\sqrt{3\over 2}\ \ \ .$$

ii)[8 pts] Show that for

$$n\ge\sqrt{2}$$

the fraction F of the surface of the cube that must be covered to prevent light from the source exiting the cube without first being totally internally reflected is

$$F~=~{\pi\over 4 (n^2-1)}\ \ \ .$$

For a refractive index $n\ge\sqrt{2}$ the circle on each face outside of which light is totally internally reflected is contained entire within the square. Therefore the fraction of the face that this circle covers is

$$F~=~{ \pi h^2\over 4 a^2}\ \ \ .$$

where h is the radius of the circle and 2a is the length of an edge of the cube. The radius of the circle is related to the angle between the edge of the circle and the normal of the face by $h~=~a\tan\theta$, and for total internal reflection $n\sin\theta~=~1$. Hence,

$$F~=~{ \pi a^2\tan^2\theta\over 4 a^2}~=~{\pi \sin^2\theta\over4\cos^2\theta}~=~{\pi\over 4\sqrt{n^2-1}}\ \ \ .$$

1B) [10 pts] A doubly convex lens is made from material with refractive index $n~=~{3\over 2}$. Its two surfaces have radii of curvature $\vert R_1\vert~=~10~{\rm m}$ and $\vert R_2\vert~=~{10\over 3}~{\rm m}$. This lens is placed a distance $d~=~5~{\rm m}$in front of a concave mirror with a radius of curvature $\vert R_m\vert~=~10~{\rm m}$. An object is placed a distance $O~=~10~{\rm m}$ in front of the lens. After first finding the image produced by the lens (which acts as an object for the mirror), find the location of the image produced by the mirror.

The focal length of the lens is

$${1\over f_l}~=~(n-1)\left( {1\over R_1}-{1\over R_2}\right)\ \ \ ,$$

with $R_1~=~10~{\rm m}$ and $R_2~=~-{10\over 3}~{\rm m}$for the doubly convex lens. With a refractive index of $n~=~{3\over 2}$, this gives $f_l~=~5~{\rm m}$. For an object located $O~=~10~{\rm m}$ from the lens, and image is produced at I, where

$${1\over O_l}+{1\over I_l}~=~{1\over f_l}\ \ \ ,$$

which gives $I_l~=~10~{\rm m}$. Since the mirror os located a distance $d~=~5~{\rm m}$ behind the lens, the image of the lens, and hence the object for the mirror is $O_m~=~-5~{\rm m}$, i.e. behind the mirror).

The radius of curvature of the lens is $\vert R_m\vert~=~10~{\rm m}$, which corresponds to a focal length of $f_m~=~5~{\rm m}$. We can find the location of the image produced by the mirror using

$${1\over O_m}+{1\over I_m}~=~{1\over f_m}\ \ \ ,$$

which gives $I_m~=~2.5~{\rm m}$, in front of the mirror. So the image is midway between the lens and the mirror.