The Workbook

1. First law: the incident ray, the reflected ray and the normal at the point of incidence lie in one plane. Second law: the angle of reflection equals the angle of incidence, \(r = i\).

2. The incident ray and the reflected ray are each \(30^\circ\) from the normal, on opposite sides. The angle between them is \(i + r = 60^\circ\).

3. Without the mirror the ray would have gone straight on. Measured from that direction, the turn is \(\delta = 180^\circ - 2i = 180^\circ - 60^\circ = 120^\circ\).

1. The image is as far behind the mirror as the object is in front. Before: \(2 + 2 = 4\) m. After the walk the person is \(1.5\) m from the mirror, so the distance is \(1.5 + 1.5 = 3\) m.

2. The image moved from \(2\) m behind the mirror to \(1.5\) m behind it: \(0.5\) m, the same as the person.

1. The eye \(E\) sees the top of the head \(H\) by a ray that reflects midway between their levels. The eye sees the feet \(F\) by a ray that reflects midway between the levels of \(E\) and \(F\). So the mirror must reach from half of \(EF\) below the eye to half of \(EH\) above the eye.

   Length \(= \dfrac{EF}{2} + \dfrac{EH}{2} = \dfrac{170}{2} + \dfrac{10}{2} = 85 + 5 = 90\) cm, half the height of the person.

2. Eye level is \(170\) cm. Lower edge: \(170 - 85 = 85\) cm above the floor. Upper edge: \(170 + 5 = 175\) cm above the floor.

1. Angles of incidence and reflection are measured from the normal. The ray is \(40^\circ\) from the mirror, so \(i = 90^\circ - 40^\circ = 50^\circ\), and \(r = i = 50^\circ\).

2. \(\delta = 180^\circ - 2i = 180^\circ - 100^\circ = 80^\circ\). Check: the ray turns through twice the angle it makes with the mirror, \(2 \times 40^\circ = 80^\circ\).

A plane mirror swaps left and right, so the clock face appears reflected about the \(12\)-\(6\) line. The actual time is \(11{:}60 - 4{:}30 = 7{:}30\).

Reason: the mirror image of the hour hand at \(4\) lies at \(8\), and the minute hand at \(6\) stays at \(6\).

\(\dfrac{360^\circ}{\theta} = \dfrac{360^\circ}{60^\circ} = 6\), an even number. The number of images is \(n = 6 - 1 = 5\) for any position of the object.

1. \(\dfrac{360^\circ}{90^\circ} = 4\) is even, so \(n = 4 - 1 = 3\) images.

2. \(O\) is at \((2, 1)\). The image in \(M_1\) (the \(x\) axis) is at \((2, -1)\). The image in \(M_2\) (the \(y\) axis) is at \((-2, 1)\). Each of these is itself reflected in the other mirror, and both give the same third image at \((-2, -1)\).

1. \(\dfrac{360^\circ}{72^\circ} = 5\), an odd number. For an object on the bisector the last two images coincide, so \(n = 5 - 1 = 4\).

2. For an object off the bisector the last two images are separate, so \(n = 5\).

\(\dfrac{360^\circ}{50^\circ} = 7.2\), not an integer. The number of images is the integer part, \(n = 7\), for every position of the object.

1. Each image in one mirror acts as an object for the other mirror. The chain never ends, so there are infinitely many images (they grow fainter along the chain).

2. \(O\) is \(1\) m in front of \(M_1\). So the nearest image in \(M_1\) is \(1\) m behind \(M_1\), at \(2\) m from \(O\). \(O\) is \(2\) m in front of \(M_2\). So the nearest image in \(M_2\) is \(2\) m behind \(M_2\), at \(4\) m from \(O\).

3. The two images lie on opposite sides of \(O\): \(2 + 4 = 6\) m apart.

1. The image is always as far behind the mirror as the object is in front. When the object comes \(1\) m closer, the image also comes \(1\) m closer. So the image moves at \(3\) m/s towards the mirror, from the other side.

2. Object and image approach each other at \(3 + 3 = 6\) m/s.

Let the mirror move a distance \(s\) towards the object. The object is now \(s\) nearer the mirror, so the image is also \(s\) nearer the mirror, that is \(2s\) nearer the object. The image moves at twice the mirror speed: \(2 \times 2 = 4\) m/s towards the object.

1. Along the normal, \(v_{\text{image}} = 2v_{\text{mirror}} - v_{\text{object}}\) (all with signs). Take the direction from the object towards the mirror as positive: \(v_{\text{object}} = +5\) m/s, \(v_{\text{mirror}} = -2\) m/s.

   \(v_{\text{image}} = 2(-2) - 5 = -9\) m/s, that is \(9\) m/s towards the object.

2. Relative velocity \(= v_{\text{image}} - v_{\text{object}} = -9 - 5 = -14\) m/s. The image approaches the object at \(14\) m/s.

The distance of the image behind the mirror does not change when the object moves parallel to the mirror. The image copies the motion: \(4\) m/s in the same direction as the object.

No. Motion of the mirror in its own plane does not change where the image is, so the image velocity stays \(4\) m/s.

1. The mirror reverses only the component along its normal (\(x\)). The components parallel to the mirror (\(y\), \(z\)) are unchanged. \(\vec v_i = (-3\hat i + 4\hat j)\) m/s, and the speed is \(\sqrt{9 + 16} = 5\) m/s, the same as the object.

2. \(\vec v_i - \vec v_o = (-3 - 3)\hat i + (4 - 4)\hat j = -6\hat i\) m/s. Object and image approach each other along the normal at \(6\) m/s.

Along the normal: \(v_{ix} = 2v_{mx} - v_{ox} = 2(2) - 3 = 1\) m/s. Parallel to the mirror the image copies the object: \(v_{iy} = 4\) m/s. So \(\vec v_i = (\hat i + 4\hat j)\) m/s.

When the mirror turns by \(\theta\), the normal also turns by \(\theta\). The angle of incidence changes by \(\theta\), so the reflected ray moves by \(\theta\) relative to the normal. The normal itself has also moved by \(\theta\). Together the reflected ray turns by \(2\theta = 20^\circ\).

1. Before: \(\delta = 180^\circ - 2(30^\circ) = 120^\circ\). After: \(\delta = 180^\circ - 2(45^\circ) = 90^\circ\).

2. The incident ray did not move, so the change in deviation is the turn of the reflected ray: \(120^\circ - 90^\circ = 30^\circ\). This is twice the \(15^\circ\) turn of the mirror.

1. \(i = 0^\circ\): \(\delta = 180^\circ\). The ray goes back along its own path.

2. \(i = 45^\circ\): \(\delta = 90^\circ\). The ray turns through a right angle.

3. \(i = 90^\circ\): \(\delta = 0^\circ\). A grazing ray is not turned at all.

1. At \(M_1\) the reflected ray makes \(90^\circ - 30^\circ = 60^\circ\) with the mirror \(M_1\). In the triangle formed by the two mirrors and this ray, the angle at \(M_2\) is \(180^\circ - 70^\circ - 60^\circ = 50^\circ\). This is the angle between the ray and \(M_2\), so \(i_2 = 90^\circ - 50^\circ = 40^\circ\).

2. At \(M_1\): \(\delta_1 = 180^\circ - 60^\circ = 120^\circ\). At \(M_2\): \(\delta_2 = 180^\circ - 80^\circ = 100^\circ\). Both turns are in the same sense, so the total is \(220^\circ\). Check: \(360^\circ - 2(70^\circ) = 220^\circ\).

1. Let the ray make angle \(\alpha\) with \(M_1\) and angle \(\beta\) with \(M_2\). In the triangle formed by the mirrors and the ray between them, \(\alpha + \beta + \theta = 180^\circ\).

   Each reflection turns the ray through twice the angle it makes with that mirror, in the same sense. Total deviation \(= 2\alpha + 2\beta = 2(180^\circ - \theta) = 360^\circ - 2\theta\). The angle of incidence does not appear.

2. \(\theta = 90^\circ\) gives \(\delta = 360^\circ - 180^\circ = 180^\circ\). A turn of \(180^\circ\) means the final ray is antiparallel to the incident ray, for any angle of incidence.

1. The reflected ray turns through twice the angle of the mirror, so its angular speed is \(2\omega = 4\) rad/s.

2. The spot is at distance \(10\) m along a ray that sweeps at \(4\) rad/s. When the ray is normal to the screen, the spot speed is \(r\,\omega_{\text{ray}} = 10 \times 4 = 40\) m/s.

\(0^\circ\). For two mirrors at \(90^\circ\) the total deviation is \(360^\circ - 2(90^\circ) = 180^\circ\), whatever the angle of incidence. Turning the pair changes the angle of incidence but not the deviation, so the final ray stays antiparallel to the incident ray.

1. The reflected ray appears to come from the image \(S'\) at \((0, -3)\). So \(S'\), \(Q\) and \(P\) lie on one line. Slope of \(S'P\): \(\dfrac{3 - (-3)}{8 - 0} = \dfrac{3}{4}\). At \(y = 0\): \(0 = -3 + \dfrac{3}{4}x\), so \(x = 4\) m.

   Check: \(S\) and \(P\) are at equal heights and \(Q\) is midway. So the ray from \(S\) and the ray to \(P\) make equal angles with the normal at \(Q\).

2. The path \(SQP\) has the same length as the straight line \(S'P\): \(\sqrt{8^2 + 6^2} = 10\) m.

1. \(f = \dfrac{R}{2} = 20\) cm. The focus of a concave mirror is in front of the mirror, against the incident light, so \(f = -20\) cm.

2. \(u = -30\) cm. \(\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = -\dfrac{1}{20} + \dfrac{1}{30} = -\dfrac{1}{60}\), so \(v = -60\) cm.

   \(m = -\dfrac{v}{u} = -\dfrac{-60}{-30} = -2\).

3. \(v\) is negative: the image is in front of the mirror, so it is real. \(m\) is negative: the image is inverted. \(|m| = 2\): the image is twice the size of the object.

\(u = -10\) cm, \(f = -20\) cm. \(\dfrac{1}{v} = -\dfrac{1}{20} + \dfrac{1}{10} = \dfrac{1}{20}\), so \(v = +20\) cm. \(m = -\dfrac{20}{-10} = +2\).

The image is behind the mirror (virtual), erect and twice the size. An object inside the focus of a concave mirror always gives this kind of image (a shaving mirror).

For a convex mirror the focus is behind the mirror: \(f = +20\) cm. \(u = -20\) cm. \(\dfrac{1}{v} = \dfrac{1}{20} + \dfrac{1}{20} = \dfrac{1}{10}\), so \(v = +10\) cm. \(m = -\dfrac{10}{-20} = +\dfrac{1}{2}\).

The image is virtual, erect and half the size of the object.

At the centre of curvature \(u = -R = -2f\) (with \(f\) negative for a concave mirror, \(u = 2f\)). \(\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{2f} = \dfrac{1}{2f}\), so \(v = 2f = u\). The image is at the centre of curvature.

\(m = -\dfrac{v}{u} = -1\): the image is real, inverted and the same size as the object.

With \(f > 0\) and \(u < 0\), \(\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{|u|}\) is a sum of two positive terms. So \(v > 0\): the image is behind the mirror, virtual.

Also \(\dfrac{1}{v} > \dfrac{1}{|u|}\), so \(v < |u|\). Then \(m = -\dfrac{v}{u} = \dfrac{v}{|u|}\) is positive and less than \(1\): erect and smaller. This holds for every object distance.

\(u = -15\) cm, \(f = -10\) cm. \(\dfrac{1}{v} = -\dfrac{1}{10} + \dfrac{1}{15} = -\dfrac{1}{30}\), so \(v = -30\) cm. \(m = -\dfrac{-30}{-15} = -2\).

Height of image \(= m \times 2 = -4\) cm: the image is \(4\) cm tall, inverted, real, \(30\) cm in front of the mirror.

A real image from a concave mirror is inverted, so \(m = -3\). \(m = -\dfrac{v}{u} = -3\) gives \(v = 3u\).

\(\dfrac{1}{3u} + \dfrac{1}{u} = \dfrac{4}{3u} = \dfrac{1}{f} = -\dfrac{1}{12}\), so \(u = -16\) cm and \(v = -48\) cm.

A convex mirror gives an erect image, so \(m = +\dfrac{1}{2}\), and \(v = -\dfrac{u}{2}\).

\(\dfrac{1}{v} + \dfrac{1}{u} = -\dfrac{2}{u} + \dfrac{1}{u} = -\dfrac{1}{u} = \dfrac{1}{15}\), so \(u = -15\) cm and \(v = +7.5\) cm.

1. \(y = \dfrac{f^2}{x} = \dfrac{400}{5} = 80\) cm. The image is \(80\) cm beyond \(F\), on the same side as the object.

2. Image distance from the pole \(= 20 + 80 = 100\) cm, so \(v = -100\) cm. Check: \(u = -25\) cm, \(\dfrac{1}{v} = -\dfrac{1}{20} + \dfrac{1}{25} = -\dfrac{1}{100}\). Correct.

1. \(u = -30\) cm, \(f = -20\) cm give \(v = -60\) cm (Question 1) and \(m = -2\).

2. \(m_L = -m^2 = -4\). The image of the rod is \(4 \times 1 = 4\) mm long, and it lies along the axis. The minus sign says that the order of the ends is reversed. The image of \(A\), the far end, is nearer the mirror than the image of \(B\).

\(u = -15\) cm, \(f = -10\) cm give \(v = -30\) cm and \(m = -2\). Each side of the square is magnified \(2\) times, so the area is magnified \(m^2 = 4\) times. Area of image \(= 4\) cm\(^2\).

1. The mirror formula is \(\dfrac{1}{v} = -\dfrac{1}{u} + \dfrac{1}{f}\). Plot \(\dfrac{1}{v}\) against \(\dfrac{1}{u}\). This is a straight line with slope \(-1\) and intercept \(\dfrac{1}{f}\) on each axis.

2. Each intercept equals \(\dfrac{1}{f} = -0.05\) cm\(^{-1}\), so \(f = -20\) cm. A negative focal length means a concave mirror.

3. \(u = v\) is the point where the line meets the line \(\dfrac{1}{v} = \dfrac{1}{u}\), that is \(\dfrac{1}{u} = \dfrac{1}{v} = -0.025\) cm\(^{-1}\). So \(u = -40\) cm, the centre of curvature.

1. Both the object and the image are in front of the mirror: \(u = -30\) cm, \(v = -15\) cm.

2. \(\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} = -\dfrac{1}{15} - \dfrac{1}{30} = -\dfrac{1}{10}\), so \(f = -10\) cm and \(R = 2f = -20\) cm.

3. \(m = -\dfrac{v}{u} = -\dfrac{-15}{-30} = -\dfrac{1}{2}\). The image is inverted and half the size, as drawn. \(v < 0\), so the image is real.

The image is on a screen, so it is real and inverted: \(m = -3\) and \(v = 3u\). Both are in front of the mirror, and the image is farther: \(|v| - |u| = 80\) cm gives \(2|u| = 80\), \(|u| = 40\) cm.

\(u = -40\) cm, \(v = -120\) cm. \(\dfrac{1}{f} = -\dfrac{1}{120} - \dfrac{1}{40} = -\dfrac{4}{120}\), so \(f = -30\) cm.

\(f = +30\) cm, \(v = +20\) cm. \(\dfrac{1}{u} = \dfrac{1}{f} - \dfrac{1}{v} = \dfrac{1}{30} - \dfrac{1}{20} = -\dfrac{1}{60}\), so \(u = -60\) cm. \(m = -\dfrac{20}{-60} = +\dfrac{1}{3}\).

\(f\) is constant, so \(\dfrac{d}{dt}\left(\dfrac{1}{v} + \dfrac{1}{u}\right) = 0\): \(-\dfrac{1}{v^2}\dfrac{dv}{dt} - \dfrac{1}{u^2}\dfrac{du}{dt} = 0\).

So \(\dfrac{dv}{dt} = -\dfrac{v^2}{u^2}\dfrac{du}{dt}\), that is \(v_I = -m^2 v_O\), since \(m^2 = \dfrac{v^2}{u^2}\). The image always moves opposite to the object along the axis, with speed \(m^2\) times the object speed.

1. \(u = -30\) cm, \(f = -20\) cm: \(v = -60\) cm, \(m = -2\). The object moves towards the mirror, so \(u\) increases towards zero: \(v_O = +2\) cm/s.

   \(v_I = -m^2 v_O = -4 \times 2 = -8\) cm/s. \(v\) becomes more negative: the image moves away from the mirror at \(8\) cm/s.

2. Faster, because \(|m| = 2 > 1\) and the image speed is \(m^2 = 4\) times the object speed.

1. \(f = +2\) m, \(u = -6\) m. \(\dfrac{1}{v} = \dfrac{1}{2} + \dfrac{1}{6} = \dfrac{2}{3}\), so \(v = +1.5\) m behind the mirror. \(m = -\dfrac{1.5}{-6} = +\dfrac{1}{4}\).

2. \(v_I = -m^2 v_O = -\dfrac{1}{16} \times 16 = -1\) m/s. \(v\) decreases from \(1.5\) m: the image moves towards the mirror at \(1\) m/s. The car seems slow and small in the mirror.

For motion perpendicular to the axis, \(h_I = m\,h_O\) with \(m\) fixed (since \(u\) does not change). Differentiating, \(v_{I\perp} = m\,v_{O\perp} = -2 \times 3 = -6\) cm/s.

The image moves at \(6\) cm/s in the direction opposite to the object, because the image is inverted.

1. \(m = -2\) as before. Along the axis: \(v_{Ix} = -m^2 v_{Ox} = -4 \times 2 = -8\) cm/s. Perpendicular: \(v_{Iy} = m\,v_{Oy} = -2 \times 3 = -6\) cm/s.

2. \(\vec v_I = (-8\hat i - 6\hat j)\) cm/s, speed \(= \sqrt{64 + 36} = 10\) cm/s.

\(|v_I| = |v_O|\) needs \(m^2 = 1\), so \(|m| = 1\) and \(|v| = |u|\). A real image the same size as the object has \(v = u\). Then \(\dfrac{2}{u} = \dfrac{1}{f}\), \(u = 2f = -40\) cm.

This is the centre of curvature. There the image is the same size as the object, and it moves at the same speed, in the opposite direction.

No. The image speed is \(m^2\) times the object speed, and \(m\) changes with position. At \(C\), \(|m| = 1\). As the object nears \(F\), \(|m|\) grows without limit, so the image speeds up and moves off to infinity.

As \(R \to \infty\), \(f \to \infty\) and \(\dfrac{1}{f} \to 0\), so \(\dfrac{1}{v} = -\dfrac{1}{u}\) and \(v = -u\). Then \(m = -\dfrac{v}{u} = +1\) and \(m^2 = 1\).

\(v_I = -m^2 v_O = -v_O\): the image moves towards the mirror as fast as the object does, from the other side. This agrees with the plane mirror result.

1. \(f = -30\) cm, \(u = -20\) cm. \(\dfrac{1}{v} = -\dfrac{1}{30} + \dfrac{1}{20} = \dfrac{1}{60}\), so \(v = +60\) cm. \(m = -\dfrac{60}{-20} = +3\). The image is virtual, erect and three times the size.

2. \(v_O = +1\) cm/s (towards the mirror). \(v_I = -m^2 v_O = -9\) cm/s. \(v\) decreases from \(+60\) cm: the virtual image moves towards the mirror at \(9\) cm/s.