A rider on a scooter glances at the round mirror on the handlebar. A car far behind looks tiny and seems hardly to gain on the scooter. A moment later it is alongside. The side mirrors of many cars carry a printed warning: objects in the mirror are closer than they appear.
Differentiate the formula
When an object moves along the axis, \(u\) changes with time, and so does \(v\). The focal length stays fixed. So the mirror formula holds at every instant, and its time derivative is zero. The rate \(du/dt\) is the velocity of the object along the axis, \(v_O\). The rate \(dv/dt\) is the velocity of the image, \(v_I\). Both are measured relative to the mirror, with the direction of the incident light positive.
\[ \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \]
Why. the mirror formula, true at every instant.
\[ -\dfrac{1}{v^2}\dfrac{dv}{dt} - \dfrac{1}{u^2}\dfrac{du}{dt} = 0 \]
Why. differentiate with respect to time; \(f\) is constant, and \(\frac{d}{dt}\frac1v = -\frac1{v^2}\frac{dv}{dt}\).
\[ \dfrac{dv}{dt} = -\dfrac{v^2}{u^2}\dfrac{du}{dt} \]
Why. solve for \(dv/dt\).
\[ v_I = -m^2\,v_O \]
Read it. \(v^2/u^2 = m^2\), since \(m = -v/u\).
The result says two things in plain words. The minus sign means that the image always moves the opposite way to the object along the axis. The factor \(m^2\) means that the image moves faster than the object where the image is larger, and slower where it is smaller. A plane mirror has \(m = +1\), so \(v_I = -v_O\), the result of the plane-mirror lesson.
An object is \(30\,\text{cm}\) in front of a concave mirror of focal length \(20\,\text{cm}\). It moves towards the mirror along the axis at \(2\,\text{cm/s}\). Find the velocity of the image.
\(u = -30\,\text{cm}\), \(f = -20\,\text{cm}\). Moving towards the mirror, \(u\) goes from \(-30\) towards zero, so it increases: \(v_O = +2\,\text{cm/s}\).
The mirror formula gives \(v = -60\,\text{cm}\), so \(m = -v/u = -2\). Then \(v_I = -m^2 v_O\).
\(v_I = -4 \times 2 = -8\,\text{cm/s}\).
\(v\) becomes more negative, so the image moves away from the mirror at \(8\,\text{cm/s}\), four times the object's speed.
At the centre of curvature \(m = -1\), so the image moves exactly as fast as the object, the other way. Between C and F the image is larger, and it moves faster. The next figure asks what happens as the object walks steadily from C towards F.
A point object moves along the axis of a concave mirror at a steady \(2\,\text{cm/s}\), from C towards F. How does its image move?
In the figure above the mirror has \(f = -20\,\text{cm}\), and the object starts at C, \(40\,\text{cm}\) away. At \(u = -30\,\text{cm}\), \(m = -2\) and the image moves at \(8\,\text{cm/s}\). At \(u = -25\,\text{cm}\), \(m = -4\) and it moves at \(32\,\text{cm/s}\). The figure stops at \(u = -23.5\,\text{cm}\), where \(v = -134.3\,\text{cm}\) and the image races at about \(65\,\text{cm/s}\). An object moving at a steady speed never gives an image moving at a steady speed.
The rule \(v_I = -m^2 v_O\) gives the speed at one instant. Over a long stretch the image does not move uniformly, so the distance it covers is not the speed times the time. For such questions, find \(v\) at the start and the end with the mirror formula and subtract. For motion relative to a moving mirror, use velocities relative to the mirror, then add the mirror's velocity back.
A candle is \(20\,\text{cm}\) in front of a concave mirror of radius of curvature \(60\,\text{cm}\). The candle moves towards the mirror at \(1\,\text{cm/s}\). Find the speed of the image, in cm/s.
An object moves along the axis of a concave mirror of focal length \(20\,\text{cm}\). At what object distance does its image move with the same speed as the object?
Moving across the axis
Now let a point object on the axis move perpendicular to it. At that instant \(u\) is not changing, so \(v\) and \(m\) are fixed too. The height of the image is \(h' = mh\), so its rate of change is \(m\) times the rate of change of \(h\). Across the axis, the image velocity is \(m\) times the object velocity, not \(m^2\).
Take the object at \(u = -30\,\text{cm}\) before a concave mirror with \(f = -20\,\text{cm}\), where \(m = -2\). If it moves up at \(3\,\text{cm/s}\), the image moves at \(-2 \times 3 = -6\,\text{cm/s}\): \(6\,\text{cm/s}\) downwards. The image is inverted, so it moves the opposite way across the axis too.
A velocity at an angle splits into two parts that follow the two rules. Give the same object a velocity \((2\hat i + 3\hat j)\,\text{cm/s}\), with \(\hat i\) along the axis towards the mirror. Along the axis \(v_{Ix} = -m^2 v_{Ox} = -4 \times 2 = -8\,\text{cm/s}\). Across it, \(v_{Iy} = m v_{Oy} = -2 \times 3 = -6\,\text{cm/s}\). The image moves at \((-8\hat i - 6\hat j)\,\text{cm/s}\), a speed of \(\sqrt{64 + 36} = 10\,\text{cm/s}\).
A point object on the axis of a concave mirror moves perpendicular to the axis at \(3\,\text{cm/s}\), at a place where the magnification is \(-2\). What is the image velocity across the axis?
The mirror on the handlebar
The rear-view mirror of a scooter is convex. A convex mirror gives an erect, smaller image of anything behind the rider, and it packs a wide view into a small round glass. Take a mirror of radius \(4\,\text{m}\), so \(f = +2\,\text{m}\) on the Cartesian convention. A car is \(6\,\text{m}\) behind, \(u = -6\,\text{m}\), approaching at \(16\,\text{m/s}\), about \(58\,\text{km/h}\).
The mirror formula gives \(1/v = 1/2 + 1/6 = 2/3\), so \(v = +1.5\,\text{m}\), behind the mirror. The magnification is \(m = -1.5/(-6) = +\tfrac14\): the car looks a quarter of its size. The car approaches, so \(v_O = +16\,\text{m/s}\). Then \(v_I = -\tfrac1{16} \times 16 = -1\,\text{m/s}\). The image creeps towards the mirror at only \(1\,\text{m/s}\).
The brain judges distance partly by size and by how fast something grows. In the mirror, a car \(6\,\text{m}\) away shows only a quarter of its height. The brain places it farther away, and it seems to crawl. As it comes close, \(m\) grows and the image speed grows as \(m^2\). The image suddenly seems to rush in, and the car is alongside. The warning printed on the glass is exact physics.
A convex mirror has a radius of curvature of \(4\,\text{m}\). A bus \(10\,\text{m}\) behind it approaches at \(12\,\text{m/s}\). Find the speed of the image of the bus, in m/s.
Explain, using the magnification, why a car seen in a convex rear-view mirror is closer than it appears.
A convex mirror forms a virtual, erect image that is smaller than the car, with \(m\) less than one. The brain judges a small image as far away, so the car seems farther than it is. The image also moves at \(m^2\) times the car's speed, so the car seems to approach slowly until it is close.
On the ghat roads of the Western Ghats, a large round convex mirror stands on a pole at each blind bend. A driver approaching the bend sees the traffic coming round the other side, small and erect, spread across a wide view. The same wide field makes the convex mirror useful at the exits of parking lots and in the corners of shops.
Now back to the scooter. At \(6\,\text{m}\) the car's image is a quarter of its size and creeps at \(1\,\text{m/s}\), though the car closes at \(16\,\text{m/s}\). At \(2\,\text{m}\) the image is half the size and moves at \(4\,\text{m/s}\). The rider sees a small, slow car that suddenly swells. A good rider turns the head before changing lanes.
- Using \(m\) instead of \(m^2\) for motion along the axis.Along the axis, \(v_I = -m^2 v_O\). The factor \(m\) belongs to motion across the axis.
- Getting the sign of \(v_O\) wrong for an object approaching the mirror.An approaching object has \(u\) increasing towards zero, so \(v_O\) is positive with the light from the left.
- Assuming a steady object gives a steady image.\(m\) changes as the object moves, so the image speeds up or slows down. Near F it races away.