A goldfish swims in a round glass bowl on a shelf. Seen through the side of the bowl, it looks plump and large. When it swims to the back of the bowl, it looks larger still. Take it out in a net, and it is a small fish.
The image of one face is the object of the next
A glass sphere, a thick lens and a fish bowl each have two curved faces. Light crosses the first face, travels through the glass or the water, and crosses the second face. Treat the two faces one at a time. The first face forms an image of the object. That image then acts as the object for the second face.
Measure the second object distance again, from the second pole \(P_2\). Let the poles be a distance \(t\) apart along the light. An image at \(v_1\) from \(P_1\) is at \(v_1 - t\) from \(P_2\). \[ u_2 = v_1 - t \] Also swap the media: the light now leaves the medium it entered at the first face.
Sometimes the first image lies beyond the second face. The light then meets the second face while it is still converging towards that image. The image is never formed, but the second face treats it as its object. Its distance \(u_2\) is positive, and it is called a virtual object. The formula needs no change; only the sign of \(u_2\) is new.
A solid glass sphere (\(\mu = 1.5\)) of radius \(10\,\text{cm}\) stands in a parallel beam of sunlight. Where does the beam come to a focus?
First face: \(\mu_1 = 1\), \(\mu_2 = 1.5\), \(u = -\infty\), \(R_1 = +10\,\text{cm}\). Second face: \(\mu_1 = 1.5\), \(\mu_2 = 1\), \(R_2 = -10\,\text{cm}\), because its centre lies behind it, against the light.
First face: \(\dfrac{1.5}{v_1} = \dfrac{0.5}{10}\), so \(v_1 = +30\,\text{cm}\) from \(P_1\). The poles are \(t = 20\,\text{cm}\) apart, so \(u_2 = 30 - 20 = +10\,\text{cm}\).
Second face: \(\dfrac{1}{v} - \dfrac{1.5}{10} = \dfrac{1 - 1.5}{-10}\), so \(\dfrac{1}{v} = 0.15 + 0.05 = 0.2\) and \(v = +5\,\text{cm}\).
The sunlight meets \(5\,\text{cm}\) beyond the far side of the sphere. For glass of \(1.5\), a sphere of any radius \(R\) focuses a beam \(R/2\) beyond its far side.
A child holds a clear glass marble, about \(1.6\,\text{cm}\) across, in the summer sun over a sheet of paper. Estimate how far beyond the marble the bright spot forms, in cm.
Not every ray through a sphere is paraxial. Trace one ray exactly, with Snell's law at both faces, through a sphere of index \(\mu = \sqrt3\). It enters at \(60°\), so \(\sin r = \sin 60°/\sqrt3 = \tfrac12\) and \(r = 30°\). Inside, the two radii to the entry and exit points are equal. So the path and the two radii form an isosceles triangle, and the ray meets the second face at \(30°\) too.
The ray in the sphere of \(\mu = \sqrt3\) meets the second face at \(30°\). At what angle does it leave, and what is its total deviation, in degrees?
In the glass-ball example, the first face forms its image \(10\,\text{cm}\) beyond the second face. Which statement is correct?
Early in the eleventh century, Ibn al-Haytham wrote a treatise on the burning sphere, a glass ball that sets fire to cloth by focusing sunlight. He traced the Sun's rays through the ball, refracting each one at both faces. He found that the rays do not all meet at one point. Rays through the outer parts of the ball cross the axis nearer to it than rays through the middle. His Book of Optics, translated into Latin, taught refraction to European scholars for centuries.
A bubble inside a glass ball
A glass paperweight often holds a small air bubble. To find where the bubble appears, follow the light from the bubble to the eye. Only one face matters, the face nearest the eye. Draw the eye on the right, so the light travels from left to right. The light starts in glass, \(\mu_1 = 1.5\), and ends in air, \(\mu_2 = 1\).
The face near the eye bulges towards the eye. Its centre lies behind it, inside the glass, on the incoming side. So for a ball of radius \(10\,\text{cm}\), \(R = -10\,\text{cm}\). Put a bubble at the centre: \(u = -10\,\text{cm}\). \[ \dfrac{1}{v} = \dfrac{1.5}{-10} + \dfrac{1 - 1.5}{-10} = -0.15 + 0.05 = -0.1 \] So \(v = -10\,\text{cm}\): the bubble appears exactly at the centre. Every ray from the centre leaves along a radius, without bending.
It does not appear the same size. The magnification is \(m = \dfrac{\mu_1 v}{\mu_2 u} = \dfrac{1.5 \times (-10)}{1 \times (-10)} = 1.5\). A bubble at the centre looks one and a half times larger and sits where it is, like the letters under the paperweight.
A glass ball (\(\mu = 1.5\)) of radius \(10\,\text{cm}\) holds a small bubble \(4\,\text{cm}\) from its centre, on a diameter AB. The bubble is nearer to A. Find how deep it appears when seen from A, and when seen from B.
From A, the bubble is \(10 - 4 = 6\,\text{cm}\) inside: \(u = -6\,\text{cm}\). From B, it is \(10 + 4 = 14\,\text{cm}\) inside: \(u = -14\,\text{cm}\). In both views \(\mu_1 = 1.5\), \(\mu_2 = 1\) and \(R = -10\,\text{cm}\).
\(\dfrac{1}{v} = \dfrac{1.5}{u} + 0.05\), since \(\dfrac{1 - 1.5}{-10} = +0.05\).
From A: \(\dfrac{1}{v} = -0.25 + 0.05 = -0.2\), so \(v = -5\,\text{cm}\). From B: \(\dfrac{1}{v} = -0.1071 + 0.05 = -0.0571\), so \(v = -17.5\,\text{cm}\).
Seen from A, the bubble appears \(5\,\text{cm}\) deep, nearer than its real \(6\,\text{cm}\). Seen from B, it appears \(17.5\,\text{cm}\) deep, farther than its real \(14\,\text{cm}\).
An air bubble in a glass sphere (\(\mu = 1.5\)) of radius \(10\,\text{cm}\) is \(4\,\text{cm}\) below the surface nearest the eye. How deep does it appear, in cm?
The fish in the bowl
A round bowl of water works like the glass ball, with water of index \(\tfrac43\) in place of glass. Ignore the thin glass wall, which bends the light very little. Let the bowl have a radius of \(15\,\text{cm}\), with the eye on the right. For light leaving the water, \(\mu_1 = \tfrac43\), \(\mu_2 = 1\) and \(R = -15\,\text{cm}\).
Put the fish at the centre, \(u = -15\,\text{cm}\). As for the bubble, every ray from the centre leaves along a radius, so \(v = -15\,\text{cm}\): the fish appears at the centre. Its magnification is \(m = \dfrac{(4/3)(-15)}{1 \times (-15)} = \tfrac43\). Slide the fish away from the centre. At \(5\,\text{cm}\) behind the near wall it appears \(4.1\,\text{cm}\) behind it, with \(m = 1.09\). At \(25\,\text{cm}\) it appears at \(32.1\,\text{cm}\), with \(m = 1.71\).
The promise is kept. The bowl acts as a weak magnifier for everything inside it, because the image is virtual and \(m = \mu_1 v/(\mu_2 u)\) comes out larger than 1. A fish at the centre looks \(\tfrac43\) times its size, in its true place. At the back wall, \(30\,\text{cm}\) behind the front, it would look twice its size.
A small fish is at the centre of a spherical bowl of water (\(\mu = \tfrac43\)) of radius \(15\,\text{cm}\). Where does it appear, and how large? Ignore the glass.
A flat surface is a sphere with an infinite radius. Let \(R \to \infty\) in the formula. The right side vanishes, and \(\dfrac{\mu_2}{v} = \dfrac{\mu_1}{u}\), so \(v = \dfrac{\mu_2}{\mu_1}\,u\). This is the apparent depth of the chapter on Refraction. A coin \(12\,\text{cm}\) under water, seen from the air, appears at \(v = \dfrac{1}{4/3}(-12) = -9\,\text{cm}\), that is, \(9\,\text{cm}\) deep.
A kingfisher hovers \(3\,\text{m}\) above the still surface of a lake. A fish \(2\,\text{m}\) below the surface looks up at it. How high above the surface does the bird appear to the fish, in m? Take \(\mu = \tfrac43\) for water.
- Using the distance from the first pole for the second face.Measure again from the second pole: \(u_2 = v_1 - t\), where \(t\) is the distance between the poles along the light.
- Keeping \(\mu_1\) and \(\mu_2\) unchanged at the second face.At the second face the light leaves the medium it entered at the first. Swap the two indices.
- Using \(m = v/u\) for a surface.For a refracting surface, \(m = \mu_1 v/(\mu_2 u)\). A fish at the centre has \(v = u\) and still looks \(\tfrac43\) times larger.