Put a glass paperweight shaped like a half ball on a newspaper, flat face down. Look straight down through the top of the dome. The letters just under the top look larger than the letters around the paperweight. Yet they do not seem lifted towards you, as a coin under water seems lifted.
One curved face between two media
The chapter on Refraction followed light across flat surfaces: a slab of glass, the surface of a pond. A flat surface bends every ray of a parallel beam by the same angle. A curved surface does not. It bends a ray near its edge more than a ray near its middle. So a curved surface can gather the rays from one point and send them to another point. It forms an image.
A spherical refracting surface is a part of a sphere that separates two transparent media. Its middle point is the pole P. The centre of the sphere is the centre of curvature C, and its radius is the radius of curvature R. The straight line through P and C is the principal axis. The end of a glass rod ground into a dome is such a surface. So is the wall of a round fish bowl, and so is the front of your eye.
To find where the image forms, every distance must carry a sign. The book uses the Cartesian sign convention, the one used by the NCERT. It is the same convention as in the chapter on Reflection, now stated for surfaces and lenses.
Draw the incident light travelling from left to right.
Three results follow at once, and they are used in every problem. A real object stands to the left of the surface, so its distance \(u\) is negative. A surface that bulges towards the incoming light has its centre on the right, so its \(R\) is positive. A surface that is hollow towards the incoming light has its centre on the left, so its \(R\) is negative.
After refraction the light goes on to the right. So a real image, where the refracted rays actually meet, lies to the right of the surface, with \(v > 0\). A virtual image, from which the refracted rays only seem to come, lies to the left, with \(v < 0\). A mirror sends the light back, so there a real image has \(v < 0\). Do not carry the mirror's habits across.
Find the sign of \(R\) from where the centre of curvature lies, never from the words "convex" or "concave". A glass surface that bulges out into the air looks convex from outside. For light leaving the glass through it, the centre lies behind, on the incoming side, so \(R < 0\).
Light travels from left to right. What is the sign of \(R\) for a surface that bulges towards the incoming light, and for one that is hollow towards it?
Deriving the refraction formula
Let a point object O sit on the principal axis in a medium of refractive index \(\mu_1\). The light meets a spherical surface of radius \(R\) and enters a medium of index \(\mu_2\). Take a ray from O that meets the surface at a point N, a small height \(h\) above the axis. The normal to the surface at N is the radius CN. The refracted ray crosses the axis at the image I.
Only rays close to the axis are used, making small angles with it. They are called paraxial rays. For them, the sine and the tangent of each angle both equal the angle in radians, and N is almost directly above P. The angles in Fig 1.2 are drawn large so that you can see them.
\[ i = \alpha + \gamma \]
Why. in the triangle ONC, the exterior angle at N equals the sum of the two opposite interior angles.
\[ \gamma = r + \beta \]
Why. in the triangle NCI, the exterior angle at C equals the sum of the two opposite interior angles.
\[ \mu_1 i = \mu_2 r \]
Why. Snell's law is \(\mu_1 \sin i = \mu_2 \sin r\), and for paraxial rays each sine equals its angle.
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Your turn. put \(i\) from step 1 and \(r\) from step 2 into step 3. Which line is it?
\[ \alpha = \dfrac{h}{-u}, \qquad \beta = \dfrac{h}{v}, \qquad \gamma = \dfrac{h}{R} \]
Why. each small angle equals its tangent, a height over a distance. The distance OP is \(-u\), because \(u\) is negative.
\[ \mu_1\left(-\dfrac{1}{u} + \dfrac{1}{R}\right) = \mu_2\left(\dfrac{1}{R} - \dfrac{1}{v}\right) \]
Why. substitute the angles and cancel \(h\). Since \(h\) has gone, every paraxial ray from O passes through the same point I.
\[ \dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R} \]
Read it. collect the terms in \(v\) and \(u\) on the left and the terms in \(R\) on the right.
The formula gives the size of the image too. Follow a ray from the top of an object of height \(h_o\) to the pole P. At P the normal is the axis itself, so the ray obeys Snell's law with small angles: \(\mu_1 h_o/(-u) = \mu_2(-h_i)/v\). Here \(h_i\) is the height of the image, negative when the image is inverted. The linear magnification \(m = h_i/h_o\) follows. \[ m = \dfrac{h_i}{h_o} = \dfrac{\mu_1 v}{\mu_2 u} \]
A long glass rod (\(\mu = 1.5\)) ends in a convex face of radius \(10\,\text{cm}\). A small object in air stands on the axis \(30\,\text{cm}\) from the pole. Find the image and its magnification.
Light goes from air into glass: \(\mu_1 = 1\), \(\mu_2 = 1.5\). The object is to the left, \(u = -30\,\text{cm}\). The face bulges towards the light, so \(R = +10\,\text{cm}\).
\(\dfrac{1.5}{v} - \dfrac{1}{-30} = \dfrac{1.5 - 1}{10}\).
\(\dfrac{1.5}{v} = 0.05 - 0.0333 = 0.0167\), so \(v = +90\,\text{cm}\). Then \(m = \dfrac{1 \times 90}{1.5 \times (-30)} = -2\).
A real image forms \(90\,\text{cm}\) inside the glass. It is inverted and twice the size of the object.
The object of the example is moved to \(10\,\text{cm}\) from the pole of the same glass face. Find the image distance \(v\), with its sign, in cm.
A point object inside glass (\(\mu = 1.5\)) is \(15\,\text{cm}\) from a surface of radius \(10\,\text{cm}\) that bulges out into the air. Light from the object leaves through that surface. Find \(v\), with its sign, in cm.
A glass surface (\(\mu = 1.5\)) of radius \(20\,\text{cm}\) is hollow towards the air. A point object in air is \(20\,\text{cm}\) from its pole. What does the surface form?
Two focal points, not one
A lens in air has one focal length, the same on both sides. A single refracting surface has two, because the two sides hold different media. The first focal point \(F_1\) is the point on the axis where an object sends its image to infinity: the rays leave the surface parallel. Put \(v \to \infty\) in the formula. \[ f_1 = -\dfrac{\mu_1 R}{\mu_2 - \mu_1} \]
The second focal point \(F_2\) is where a parallel beam, from a very distant object, comes to a focus. Put \(u \to -\infty\). \[ f_2 = \dfrac{\mu_2 R}{\mu_2 - \mu_1} \] For the glass rod of the example, \(f_1 = -\dfrac{1 \times 10}{0.5} = -20\,\text{cm}\) and \(f_2 = \dfrac{1.5 \times 10}{0.5} = +30\,\text{cm}\). The first focus is \(20\,\text{cm}\) out in the air, and the second is \(30\,\text{cm}\) inside the glass.
Divide one focal length by the other: \(f_2/f_1 = -\mu_2/\mu_1\). The minus sign says that the two foci lie on opposite sides of the surface. The sizes are in the ratio of the refractive indices: the focus inside the denser medium is the farther one. Both focal lengths also give the same number, the power of the surface. \[ P = \dfrac{\mu_2}{f_2} = -\dfrac{\mu_1}{f_1} = \dfrac{\mu_2 - \mu_1}{R} \] With \(R\) in metres, \(P\) is in dioptres. For the rod, \(P = 0.5/0.1 = 5\,\text{D}\).
Many problem books quote the two focal lengths of a surface as sizes: "20 cm and 30 cm", with \(f_2/f_1 = \mu_2/\mu_1\). This book keeps the signs, \(f_1 = -20\,\text{cm}\) and \(f_2 = +30\,\text{cm}\). Both say the same thing. Only the signed values can go straight into the formula.
A long glass rod (\(\mu = 1.5\)) has a hemispherical end of radius \(5\,\text{cm}\). A point object in air is \(20\,\text{cm}\) from the end. Find \(v\), with its sign, in cm.
For light going from air into glass (\(\mu = 1.5\)) through a convex surface, how do the two focal lengths compare?
Light entering your eye crosses one curved surface first: the front of the cornea. Its radius is about \(7.8\,\text{mm}\), and the index of the cornea is about \(1.376\). Its power as a surface is \((1.376 - 1)/0.0078 \approx 48\,\text{D}\). That is more than the lens inside the eye provides. This is why swimmers see a blur under water: water of index \(1.33\) almost cancels the bending at the cornea.
Now return to the paperweight. The flat face lies on the paper, and the centre of the dome is the middle of that flat face. So a letter right under the top sits at the centre of curvature C. Light goes up from the letter into the air: \(\mu_1 = 1.5\), \(\mu_2 = 1\), and \(R = u\) (for a dome of radius \(3\,\text{cm}\), both are \(-3\,\text{cm}\)). The formula gives \(v = u\): the image is exactly where the letter is. Every ray from C meets the dome along a radius and passes straight out.
The letter still looks larger. The rays from the edges of a letter are not quite on radii, and the magnification counts them: \(m = \dfrac{\mu_1 v}{\mu_2 u} = \dfrac{1.5\,u}{1 \times u} = 1.5\). The letters under the dome are one and a half times their real size, and they stay on the paper.
- Taking \(R\) as positive because the surface "is convex".Find where the centre of curvature lies. Along the incident light, \(R > 0\); against it, \(R < 0\). Decide for each surface in each problem.
- Putting \(\mu_1\) for glass whenever glass appears.\(\mu_1\) is the medium the light comes from, and \(\mu_2\) the medium it enters. When light leaves glass, \(\mu_1\) is the glass.
- Reading \(v > 0\) as a virtual image, as for a mirror.After refraction the light goes on, so a real image lies along the light, with \(v > 0\). A virtual image has \(v < 0\).