The Workbook

\(\dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R}\). \(u\), \(v\) and \(R\) are measured from the pole, positive in the direction of the incident light. \(\mu_1\) is the index of the medium the light comes from and \(\mu_2\) that of the medium it enters. \(R\) is the distance to the centre of curvature, with its sign.

With \(R \to \infty\) the right side is zero: \(\dfrac{\mu_2}{v} = \dfrac{\mu_1}{u}\), so \(v = \dfrac{\mu_2}{\mu_1}u\). Coin: \(\mu_1 = \dfrac{4}{3}\), \(\mu_2 = 1\), \(u = -12\): \(v = \dfrac{1}{4/3}(-12) = -9\) cm. It appears \(9\) cm deep.

\(\mu_1 = 1\), \(\mu_2 = 1.5\), \(u = -30\) cm, \(R = +10\) cm. \(\dfrac{1.5}{v} + \dfrac{1}{30} = \dfrac{0.5}{10} = 0.05\), so \(\dfrac{1.5}{v} = 0.05 - 0.0333 = 0.0167\) and \(v = 90\) cm. A real image \(90\) cm inside the glass.

\(\dfrac{1.5}{v} + \dfrac{1}{10} = 0.05\), so \(\dfrac{1.5}{v} = -0.05\) and \(v = -30\) cm. The image is virtual, \(30\) cm in front of the surface on the object's side.

Image at infinity: \(-\dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R}\), so \(f_1 = \dfrac{\mu_1R}{\mu_2 - \mu_1} = \dfrac{1 \times 10}{0.5} = 20\) cm. Object at infinity: \(f_2 = \dfrac{\mu_2R}{\mu_2 - \mu_1} = \dfrac{1.5 \times 10}{0.5} = 30\) cm. Light bends by the same amount, but it travels in different media on the two sides. So the two focal lengths are in the ratio \(\mu_1 : \mu_2\).

\(m = \dfrac{\mu_1v}{\mu_2u} = \dfrac{1 \times 90}{1.5 \times (-30)} = -2\). The image is real, inverted and twice the size of the object.

1. \(\dfrac{1.5}{v} + \dfrac{1}{20} = 0.05\), so \(\dfrac{1.5}{v} = 0\): \(v = \infty\). The refracted rays are parallel.

2. \(20\) cm is the first focal length of this surface (Q5). An object at the first focus gives a parallel beam inside the glass.

1. First surface, \(u = -\infty\), \(R = +10\): \(\dfrac{1.5}{v} = \dfrac{0.5}{10}\), so \(v = 30\) cm from the first pole, that is \(10\) cm beyond the far surface.

2. For the second surface this image is a virtual object at \(u = +10\) cm. Now \(\mu_1 = 1.5\), \(\mu_2 = 1\), \(R = -10\): \(\dfrac{1}{v} - \dfrac{1.5}{10} = \dfrac{-0.5}{-10} = 0.05\), so \(\dfrac{1}{v} = 0.2\) and \(v = 5\) cm. The beam focuses \(5\) cm beyond the sphere.

1. Light goes from the bubble leftwards to the eye: \(\mu_1 = 1.5\), \(\mu_2 = 1\). \(u = -3\) cm. The centre \(C\) lies on the far side from the light's direction: \(R = -6\) cm.

2. \(\dfrac{1}{v} + \dfrac{1.5}{3} = \dfrac{1 - 1.5}{-6} = 0.083\), so \(\dfrac{1}{v} = 0.083 - 0.5 = -0.417\) and \(v = -2.4\) cm. The bubble appears \(2.4\) cm below the surface.

1. \(u = -10\), \(R = -10\), \(\mu_1 = 1.5\), \(\mu_2 = 1\): \(\dfrac{1}{v} + \dfrac{1.5}{10} = \dfrac{-0.5}{-10} = 0.05\), so \(\dfrac{1}{v} = -0.1\) and \(v = -10\) cm: the image is at \(O\) itself.

2. \(O\) is the centre of curvature of the curved face. Every ray from \(O\) meets the surface along a radius, at normal incidence, and passes through undeviated.

\(\mu_1 = 1\), \(\mu_2 = 1.5\), \(u = -20\), \(R = -20\): \(\dfrac{1.5}{v} + \dfrac{1}{20} = \dfrac{0.5}{-20} = -0.025\), so \(\dfrac{1.5}{v} = -0.075\) and \(v = -20\) cm. Virtual image \(20\) cm in front. \(m = \dfrac{\mu_1v}{\mu_2u} = \dfrac{-20}{1.5 \times (-20)} = \dfrac{2}{3}\): upright and smaller.

Light goes from glass to air: \(\mu_1 = 1.5\), \(\mu_2 = 1\), \(u = -15\). The surface bulges towards the air, so its centre is inside the glass, on the incoming side: \(R = -10\). \(\dfrac{1}{v} + \dfrac{1.5}{15} = \dfrac{-0.5}{-10} = 0.05\), so \(\dfrac{1}{v} = -0.05\) and \(v = -20\) cm. A virtual image \(20\) cm inside the glass.

At the centre. Every ray from the centre travels along a radius and strikes the surface at normal incidence, so it is not bent. The rays appear to come from the centre, exactly where the object is.

\(\mu_1 = 1\), \(\mu_2 = 1.5\), \(u = -20\), \(R = +5\): \(\dfrac{1.5}{v} + \dfrac{1}{20} = \dfrac{0.5}{5} = 0.1\), so \(\dfrac{1.5}{v} = 0.05\) and \(v = 30\) cm inside the rod.

The fish is inside water behind a surface that is convex towards the viewer. For an object inside such a surface the image is virtual. Here \(m = \dfrac{\mu_1v}{\mu_2u}\) with \(\mu_1 > \mu_2\) and \(|v|\) close to \(|u|\), so \(m\) comes out larger than \(1\). The bowl acts as a weak magnifier.

Rays from the centre leave along radii, so the fish appears at the centre: \(v = u = -15\) cm. \(m = \dfrac{\mu_1v}{\mu_2u} = \dfrac{(4/3)(-15)}{1 \times (-15)} = \dfrac{4}{3}\): it looks \(1.33\) times larger.

1. From \(A\) the bubble is \(6\) cm inside: \(u = -6\), \(R = -10\). \(\dfrac{1}{v} + \dfrac{1.5}{6} = 0.05\), so \(\dfrac{1}{v} = -0.2\) and \(v = -5\) cm: apparent depth \(5\) cm.

2. From \(B\) it is \(14\) cm inside: \(\dfrac{1}{v} + \dfrac{1.5}{14} = 0.05\), so \(\dfrac{1}{v} = -0.057\) and \(v = -17.5\) cm: apparent depth \(17.5\) cm.

\(u = -10\), \(R = -10\): \(\dfrac{1}{v} + 0.15 = 0.05\), so \(v = -10\) cm: it appears at the centre, where it is. \(m = \dfrac{1.5 \times (-10)}{1 \times (-10)} = 1.5\): it looks one and a half times its size.

Plane surface: \(v = \dfrac{\mu_2}{\mu_1}u\). Fish seen from air: \(\dfrac{1}{4/3} \times 2 = 1.5\) m deep. Bird seen from water: \(\dfrac{4/3}{1} \times 3 = 4\) m high.

With light going left to right, a surface convex towards the incoming light has its centre of curvature to the right. That is the positive direction: \(R > 0\). A surface concave towards the incoming light has its centre on the left: \(R < 0\).

\(P = \dfrac{\mu_2 - \mu_1}{R} = \dfrac{1.5 - 1}{0.1} = 5\) D.

With \(R = \infty\): \(\dfrac{\mu_2}{v} = \dfrac{\mu_1}{u}\), so \(v = \dfrac{\mu_2}{\mu_1}u\). The image distance, the apparent depth, is the real depth \(|u|\) multiplied by \(\dfrac{\mu_2}{\mu_1}\), and it lies on the same side as the object.

\(f_1 = 20\) cm: an object here gives an image at infinity. \(f_2 = 30\) cm: an object at infinity gives an image here, inside the glass.

Set \(v = \infty\): \(f_1 = -u = \dfrac{\mu_1R}{\mu_2 - \mu_1}\). Set \(u = -\infty\): \(f_2 = v = \dfrac{\mu_2R}{\mu_2 - \mu_1}\). Dividing, \(\dfrac{f_2}{f_1} = \dfrac{\mu_2}{\mu_1}\): the focal length on each side is proportional to the refractive index on that side.

\(\dfrac{1}{f} = (\mu - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)\). \(R_1\) is the radius of the first face met by the light and \(R_2\) that of the second. Each is positive if its centre lies on the far side of the lens along the light. Equiconvex: \(R_1 = +20\), \(R_2 = -20\): \(\dfrac{1}{f} = 0.5\left(\dfrac{1}{20} + \dfrac{1}{20}\right) = 0.05\), so \(f = 20\) cm.

1. \(R_1 = +10\) cm, \(R_2 = \infty\): \(\dfrac{1}{f} = 0.5 \times \dfrac{1}{10} = 0.05\), so \(f = 20\) cm.

2. No. Turned round, \(R_1 = \infty\) and \(R_2 = -10\): \(\dfrac{1}{f} = 0.5\left(0 + \dfrac{1}{10}\right)\), the same \(20\) cm. A thin lens has the same focal length from either side.

Biconcave: \(R_1 = -15\), \(R_2 = +15\). \(\dfrac{1}{f} = 0.5\left(-\dfrac{1}{15} - \dfrac{1}{15}\right) = -\dfrac{1}{15}\), so \(f = -15\) cm: diverging.

\(\dfrac{1}{f} = 0.5\left(\dfrac{1}{10} - \dfrac{1}{20}\right) = 0.5 \times 0.05 = 0.025\), so \(f = 40\) cm. Positive: converging, because the more strongly curved face is the convex one.

First face, air to glass, object at \(u\): \(\dfrac{\mu}{v_1} - \dfrac{1}{u} = \dfrac{\mu - 1}{R_1}\). Second face, glass to air, with the first image as object (the lens is thin, so distances are unchanged): \(\dfrac{1}{v} - \dfrac{\mu}{v_1} = \dfrac{1 - \mu}{R_2}\). Adding: \(\dfrac{1}{v} - \dfrac{1}{u} = (\mu - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)\). With \(u = -\infty\), \(v = f\): the lens maker's formula.

\(\dfrac{1}{30} = (\mu - 1)\left(\dfrac{1}{30} + \dfrac{1}{30}\right) = (\mu - 1)\dfrac{2}{30}\), so \(\mu - 1 = 0.5\) and \(\mu = 1.5\).

In water the relative index is \(\dfrac{1.5}{4/3} = 1.125\). \(\dfrac{1}{f'} = 0.125 \times 0.1 = 0.0125\), so \(f' = 80\) cm: four times the air value.

1. Refraction depends on the ratio of the indices of glass and its surroundings. In water that ratio is \(1.125\) instead of \(1.5\), so each face bends the light much less and the focus moves further away.

2. \(\dfrac{f'}{f} = \dfrac{\mu_g - 1}{\mu_g/\mu_m - 1} = \dfrac{0.5}{0.125} = 4\), so \(f' = 80\) cm.

The relative index is \(1\), so \(\mu_{\text{rel}} - 1 = 0\) and \(\dfrac{1}{f} = 0\): \(f \to \infty\). Light passes without bending, and since there is no refraction or reflection at its faces the lens becomes invisible in the liquid.

\(\mu_{\text{rel}} = \dfrac{1.5}{1.6} = 0.9375\). \(\dfrac{1}{f} = (0.9375 - 1) \times 0.1 = -0.00625\), so \(f = -160\) cm. The convex lens now diverges light, because it is optically thinner than its surroundings.

\(\mu_{\text{rel}} = \dfrac{1}{4/3} = 0.75\). \(\dfrac{1}{f} = (0.75 - 1)\left(\dfrac{1}{10} + \dfrac{1}{10}\right) = -0.05\), so \(f = -20\) cm. A convex bubble of air in water diverges light, the opposite of a glass lens in air.

Power \(P = \dfrac{1}{f}\) with \(f\) in metres, measured in dioptres (D). In air: \(\dfrac{1}{0.2} = 5\) D. In water: \(\dfrac{1}{0.8} = 1.25\) D.

In air \(\dfrac{1}{f} = (\mu_g - 1)K\) where \(K = \dfrac{1}{R_1} - \dfrac{1}{R_2}\). In the medium \(\dfrac{1}{f'} = \left(\dfrac{\mu_g}{\mu_m} - 1\right)K\). Dividing, \(\dfrac{f'}{f} = \dfrac{\mu_g - 1}{\mu_g/\mu_m - 1}\), so \(f' = f\dfrac{\mu_g - 1}{\mu_g/\mu_m - 1}\).

Spherical aberration is the failure of a lens with spherical faces to bring all parallel rays to one focus. Rays through the edge focus nearer than rays near the axis, so a point object gives a blurred spot. It occurs because a sphere is not the ideal shape for focusing. It is reduced by a stop that admits only paraxial rays. It is also reduced by a plano-convex lens with its curved face towards the parallel beam, or by combining lenses so that their aberrations cancel.

1. The marginal rays, at \(F_m\). They meet the faces at larger angles and are bent more than the thin-lens formula allows.

2. A stop blocks the marginal rays, leaving only the paraxial ones, so the image becomes sharper, at the cost of brightness.

\(\dfrac{1}{f} = (\mu - 1)\dfrac{2}{20} = \dfrac{\mu - 1}{10}\). Violet: \(f_V = \dfrac{10}{0.55} \approx 18.2\) cm. Red: \(f_R = \dfrac{10}{0.5} = 20\) cm. Longitudinal chromatic aberration \(= f_R - f_V \approx 1.8\) cm.

\(\omega = \dfrac{\mu_V - \mu_R}{\mu_Y - 1}\), the ratio of the spread of the index to its mean deviation. \(\omega = \dfrac{1.55 - 1.5}{1.525 - 1} = \dfrac{0.05}{0.525} \approx 0.095\).

1. \(\dfrac{0.02}{20} + \dfrac{0.04}{f_2} = 0\), so \(f_2 = -40\) cm: a diverging flint lens.

2. \(\dfrac{1}{F} = \dfrac{1}{20} - \dfrac{1}{40} = \dfrac{1}{40}\), so \(F = 40\) cm, converging and the same for all colours.

The condition \(\dfrac{\omega_1}{f_1} = -\dfrac{\omega_2}{f_2}\) with \(\omega_1 = \omega_2\) would need \(f_2 = -f_1\), and then \(\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2} = 0\): no lens at all. With a flint of higher dispersive power, a weaker concave lens cancels the colour spread while leaving net converging power.

\(f_R - f_V \approx \omega f = 0.02 \times 20 = 0.4\) cm.

1. Violet. Its refractive index is larger, so it is bent more and \(f_V < f_R\).

2. At \(F_R\) the red light is a point but the violet has already crossed and spread out. One sees a red centre with a bluish-violet ring around it.

\(\dfrac{f'}{f} = \dfrac{\mu_g - 1}{\mu_g/\mu_m - 1} = \dfrac{0.5}{1.2 - 1} = 2.5\), so \(f' = 50\) cm.

\(\dfrac{1}{30} = 0.5 \times \dfrac{1}{R}\), so \(R = 15\) cm.

Equiconvex: \(R_1 = R\), \(R_2 = -R\). \(\dfrac{1}{f} = (1.5 - 1)\left(\dfrac{1}{R} + \dfrac{1}{R}\right) = 0.5 \times \dfrac{2}{R} = \dfrac{1}{R}\), so \(f = R\).

\(\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2}\), so powers add: \(P = P_1 + P_2\). \(\dfrac{1}{F} = \dfrac{1}{20} - \dfrac{1}{30} = \dfrac{1}{60}\), \(F = 60\) cm. \(P = 5 - 3.33 = 1.67\) D.

1. \(\dfrac{1}{F} = \dfrac{1}{20} + \dfrac{1}{30} - \dfrac{10}{600} = 0.05 + 0.0333 - 0.0167 = 0.0667\), so \(F = 15\) cm.

2. \(d = 0\): \(\dfrac{1}{F} = \dfrac{1}{20} + \dfrac{1}{30} = \dfrac{1}{12}\), \(F = 12\) cm. Separating the lenses weakens the combination.

\(P_1 = 5\) D, \(P_2 = 3.33\) D, \(d = 0.1\) m: \(P = 5 + 3.33 - 0.1 \times 16.67 = 6.67\) D. \(F = \dfrac{1}{6.67} = 0.15\) m \(= 15\) cm, as in Q2.

\(P = \dfrac{1}{0.2} - \dfrac{1}{0.2} = 0\): the pair has no focusing power. Light passes through as through a flat slab of glass, with \(F = \infty\).

\(\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{F}\) with \(u = -30\), \(F = 60\): \(\dfrac{1}{v} = \dfrac{1}{60} - \dfrac{1}{30} = -\dfrac{1}{60}\), so \(v = -60\) cm: a virtual, upright image on the object's side, twice the size.

\(\dfrac{1}{12} = \dfrac{1}{20} + \dfrac{1}{f_2}\), so \(\dfrac{1}{f_2} = \dfrac{5 - 3}{60} = \dfrac{1}{30}\) and \(f_2 = 30\) cm.

1. Each half still has both curved faces with the same radii, so the lens maker's formula gives the same \(f = 20\) cm. Only the aperture is halved.

2. Each piece is plano-convex with one face of radius \(20\) cm: \(\dfrac{1}{f} = 0.5 \times \dfrac{1}{20}\), so \(f = 40\) cm, double the original.

The focal length depends only on the curvatures of the two faces. Cut \(A\) leaves each piece with both original faces, so nothing in the formula changes. Cut \(B\) replaces one curved face by a flat one, removing half the bending power.

Two lenses of \(40\) cm in contact: \(\dfrac{1}{F} = \dfrac{1}{40} + \dfrac{1}{40}\), \(F = 20\) cm, the original lens. With the flat faces outward the two curved faces meet in the middle. For thin lenses in contact only the two powers add, so the result is again \(20\) cm.

Two images. Each half is a complete lens with its own optical centre and principal axis, and the two axes are now slightly displaced. A single point object gives one image from each half (Billet's split lens).

The image is only half as bright. Half of the lens area is gone, so half the light from the object is gathered. The image is still complete, because every part of a lens forms the whole image.

\(P_{\text{eq}} = 2P_L + P_M\). A silvered flat face is a plane mirror with \(P_M = 0\): \(P_{\text{eq}} = \dfrac{2}{20} = \dfrac{1}{10}\) per cm, so \(F_{\text{eq}} = 10\) cm. The system behaves as a concave mirror of focal length \(10\) cm.

1. \(P_L = \dfrac{1}{f} = \dfrac{1}{20}\) cm\(^{-1}\). The silvered curved face is a concave mirror of radius \(10\) cm, focal length \(5\) cm: \(P_M = \dfrac{1}{5}\) cm\(^{-1}\).

2. \(P_{\text{eq}} = \dfrac{2}{20} + \dfrac{1}{5} = \dfrac{3}{10}\) cm\(^{-1}\), so \(F_{\text{eq}} = \dfrac{10}{3} \approx 3.3\) cm, a concave mirror.

\(P_L = \dfrac{1}{20}\); the silvered face has radius \(20\) cm, so \(P_M = \dfrac{2}{20} = \dfrac{1}{10}\). \(P_{\text{eq}} = \dfrac{2}{20} + \dfrac{1}{10} = \dfrac{1}{5}\) cm\(^{-1}\), so \(F_{\text{eq}} = 5\) cm.

\(P_L = -\dfrac{1}{20}\), \(P_M = 0\): \(P_{\text{eq}} = -\dfrac{2}{20} = -\dfrac{1}{10}\) cm\(^{-1}\), so \(F_{\text{eq}} = -10\) cm. It behaves as a convex mirror of focal length \(10\) cm.

Concave mirror, \(f = -10\) cm in the mirror convention, \(u = -15\): \(\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = -\dfrac{1}{10} + \dfrac{1}{15} = -\dfrac{1}{30}\), so \(v = -30\) cm. A real, inverted image \(30\) cm in front, twice the size.

The light is refracted by the lens on its way in and reflected by the silvered surface. It is then refracted by the lens again on its way out. Each passage through the lens adds its power \(P_L\), and the mirror adds \(P_M\), so the total is \(2P_L + P_M\).

1. \(f = \dfrac{100^2 - 40^2}{4 \times 100} = \dfrac{8400}{400} = 21\) cm.

2. \(u + v = 100\) and \(v - u = 40\): \(u = 30\) cm at \(L_1\), and by symmetry \(u = 70\) cm at \(L_2\).

\(O = \sqrt{I_1I_2} = \sqrt{4 \times 1} = 2\) cm. \(m_1 = \dfrac{4}{2} = 2\) and \(m_2 = \dfrac{1}{2}\). Their product is \(1\), as it must be for the two conjugate positions.

\(D \ge 4f\), so \(D_{\min} = 84\) cm. With \(u + v = D\) and \(\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}\), the sum \(u + v\) is least when \(u = v = 2f\), giving \(D = 4f\). For smaller \(D\) no real image can fall on the screen and \(d^2 = D^2 - 4fD\) would be negative.

For a real image, \(u + v = D\) (magnitudes). The two lens positions are \(d\) apart, so \(v - u = d\) at the first position: \(u = \dfrac{D - d}{2}\), \(v = \dfrac{D + d}{2}\). Then \(\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v} = \dfrac{2}{D - d} + \dfrac{2}{D + d} = \dfrac{4D}{D^2 - d^2}\), so \(f = \dfrac{D^2 - d^2}{4D}\).

1. \(u_1 = 30\) cm, \(v_1 = 70\) cm; \(u_2 = 70\) cm, \(v_2 = 30\) cm.

2. The lens formula is symmetric in \(u\) and \(v\). If an object at \(30\) cm images at \(70\) cm, an object at \(70\) cm images at \(30\) cm. The second position simply swaps the roles of the two distances.

Only the displacement \(d\) of the lens and the fixed distance \(D\) need be measured. Neither requires knowing where the optical centre of the lens lies, which is hard to locate for a thick lens in a mount.

At the first position \(m_1 = \dfrac{I_1}{O} = \dfrac{v}{u}\) and at the second \(m_2 = \dfrac{I_2}{O} = \dfrac{u}{v}\). Multiplying, \(\dfrac{I_1I_2}{O^2} = 1\), so \(O = \sqrt{I_1I_2}\).