Coaching Centre
Images in mirrors and lenses
The ray model of light, how plane and curved mirrors and converging and diverging lenses form images, the standard rays for ray diagrams, real and virtual images, magnification and the thin lens equation.
A phone camera lens is a few millimetres across, yet it throws a sharp, upside-down image of a whole room onto a sensor smaller than your fingernail. A dentist's mirror shows an enlarged upright view of the back of a tooth, and a shop's security mirror shows the whole aisle shrunk into one curved surface. Each of these images can be predicted with a ruler and two or three straight lines.
In Year 9 you used a ray box to see light reflect from mirrors and bend through lenses. Here you will locate images exactly, describe them in the HSC terms (position, real or virtual, upright or inverted, size) and calculate them.
The ray model
For objects much bigger than the wavelength of light (about 500 nm), we can ignore diffraction and treat light as travelling in straight lines, called rays, which change direction only at a reflecting or refracting surface. Every point on an object sends rays out in all directions; an image forms where rays from one object point meet again, or appear to come from.
- A real image forms where rays actually meet. It can be caught on a screen. Real images from a single mirror or lens are inverted.
- A virtual image forms where rays only appear to come from when traced backwards (dashed lines). It cannot be caught on a screen. Virtual images from a single mirror or lens are upright.
Magnification compares image height with object height:
$M = \dfrac{h_i}{h_o}$
Mirrors
A plane mirror forms a virtual, upright image the same size as the object ($M = 1$), as far behind the mirror as the object is in front, and laterally inverted (left and right swapped).
A concave (converging) mirror curves inwards like the inside of a spoon. Rays parallel to the principal axis reflect through the focal point F, halfway between the mirror and the centre of curvature C. A convex (diverging) mirror bulges outwards; parallel rays reflect so they appear to come from a focal point behind it. Use any two of these rays for a concave mirror:
- A ray parallel to the principal axis reflects through F.
- A ray through F reflects parallel to the axis.
- A ray through C reflects straight back along itself.
- A ray hitting the centre of the mirror reflects at an equal angle on the other side of the axis.
Lenses
A converging (convex) lens is thicker in the middle and refracts parallel rays through a focal point F on the far side, at the focal length $f$ from the centre. A diverging (concave) lens is thinner in the middle and spreads parallel rays out so they appear to come from F on the near side. A lens has a focal point on each side. The standard rays are:
- A ray parallel to the axis refracts through F on the far side (converging) or away as if from F on the near side (diverging).
- A ray through the optical centre passes straight through.
- A ray through F on the near side (or heading for F on the far side, for a diverging lens) leaves parallel to the axis.
| Object position (converging lens or concave mirror) | Image | Use |
|---|---|---|
| Beyond 2F (or C) | real, inverted, diminished | camera, eye |
| At 2F (or C) | real, inverted, same size | photocopier |
| Between F and 2F | real, inverted, enlarged | projector |
| At F | no image (rays leave parallel) | torch, lighthouse |
| Inside F | virtual, upright, enlarged | magnifying glass, make-up mirror |
Diverging lenses and convex mirrors always give a virtual, upright, diminished image, which is why convex mirrors are used for wide-angle views in shops and on car side mirrors.
The thin lens equation
Similar triangles in a ray diagram give a relationship between the object distance $u$, the image distance $v$ and the focal length $f$:
$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}, \qquad M = -\dfrac{v}{u}$
These notes use the "real is positive" convention: distances to real objects and images are positive, a virtual image has a negative $v$, and diverging lenses and convex mirrors have a negative $f$. The same equation works for curved mirrors. With this convention a negative $M$ means inverted and a positive $M$ means upright. Some textbooks state other sign conventions; whichever one your teacher uses, check your answer against a ray diagram.
An object 3.0 cm high is 15 cm from a converging lens of focal length 10 cm. Find the image position, magnification and height, and describe the image.
A stamp is 5.0 cm from a converging lens of focal length 10 cm. Find the image.
- At the start (converging lens, u = 30 cm, f = 10 cm) the readouts show v = 15.0 cm and M = −0.50: real, inverted, diminished.
- Move the object to u = 15 cm: v = 30.0 cm and M = −2.00. Now u = 10 cm: what happens to the rays?
- Set u = 5 cm: v = −10.0 cm, M = 2.00. Follow the dashed lines back to the virtual image.
- Choose Diverging lens with u = 30 cm: v = −7.5 cm and M = 0.25. Try every object distance: can you ever get a real image?
- Choose Concave mirror with u = 30 cm and f = 10 cm. Compare v with the converging lens.
Test yourself
An object is 40 cm from a converging lens of focal length 10 cm. The image is
$1/v = 1/10 - 1/40 = 3/40$, $v = 13.3$ cm, $M = -0.33$: real, inverted, smaller.
Which device gives only virtual, upright, diminished images?
Convex mirrors diverge light, so the image is always virtual, upright and smaller. A plane mirror gives the same size.
A ray parallel to the principal axis strikes a converging lens. After the lens it
That is the definition of the focal point of a converging lens.
A real image is 24 cm from a lens and the object is 8.0 cm from it. The magnification is
$M = -v/u = -24/8.0 = -3.0$: inverted and three times the size.
Label the lens ray diagram.
Real or virtual?
Rays parallel to the axis of a converging lens pass through the point. An image that can be caught on a screen is a image, and from a single lens it is . An object closer to a converging lens than F gives a image. A negative image distance in the thin lens equation means the image is .
An object 2.0 cm tall is placed 12 cm in front of a converging lens of focal length 8.0 cm. (a) Calculate the position and height of the image. (b) Describe the image fully. (c) Explain how a ray diagram drawn to scale would confirm your answer.
(a) $\dfrac{1}{v} = \dfrac{1}{8.0} - \dfrac{1}{12} = \dfrac{3 - 2}{24} = \dfrac{1}{24}$, so $v = 24$ cm. $M = -v/u = -24/12 = -2.0$, so $h_i = -2.0 \times 2.0 = -4.0$ cm.
(b) The image is real, inverted, enlarged (twice the size, 4.0 cm tall) and 24 cm from the lens on the opposite side to the object.
(c) Draw the lens with F marked 8.0 cm on each side and the object 12 cm away. A ray from the top of the object parallel to the axis refracts through F on the far side; a ray from the top through the optical centre goes straight on. Where they cross is the top of the image. To scale, they cross 24 cm beyond the lens and 4.0 cm below the axis, confirming the calculation.