Ray Optics and Optical Instruments

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XII, Chapter 9.

Why does a swimming pool look shallower than it is?

Stand at the edge of a pool and the bottom seems to be about three quarters of its real depth. A straw in a glass of water looks bent. A diamond throws back far more sparkle than a piece of glass cut the same way. All three come from one fact: light changes speed when it moves between materials, and that bends its path.

In this chapter we use the ray picture. Visible light has a wavelength of about half a micrometre, far smaller than mirrors, lenses and everyday objects, so we can treat light as straight lines that bend only at surfaces. This is the ray approximation, and it works until slits and obstacles become comparable to the wavelength, which is the story of the next chapter.

Reflection and spherical mirrors

The laws of reflection are two statements. The angle of reflection equals the angle of incidence, and the incident ray, the reflected ray and the normal at the point of incidence all lie in one plane. On a curved mirror the normal at any point is the radius drawn to that point.

A spherical mirror is a piece of a sphere. Its important points are the pole P (the centre of the mirror), the centre of curvature C and the principal focus F, where rays parallel to the axis meet (or appear to come from). The distance PC is the radius of curvature R, and the distance PF is the focal length f. For rays close to the axis (paraxial rays):

$$f = R/2$$

Why half? A ray parallel to the axis hits the mirror at some point and reflects at an angle equal to the angle it makes with the radius from C. For a small angle, the reflected ray crosses the axis at a point that turns out to be midway between P and C.

One sign convention for everything. Pick the pole (or optical centre) as the origin. Distances measured in the direction of the incoming light are positive, distances opposite to it are negative, and heights above the principal axis are positive.

With this convention, one equation and one magnification formula cover every case:

$$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}, \qquad m = \dfrac{h'}{h} = -\dfrac{v}{u}$$

A negative m means the image is inverted (and real). A positive m means erect (and virtual). The size of m tells you how much it is magnified.

A convex mirror always gives a virtual, erect and diminished image between P and F, whatever the object distance. That wide field of view is why it is used as a rear view mirror.

Worked example: concave mirror. An object stands 10 cm in front of a concave mirror of radius of curvature 15 cm. Find the image position and magnification.

The image is 30 cm in front of the mirror, real, inverted and three times the size of the object.

Common wrong step. Using $f = +7.5 \text{ cm}$ because "R is a positive length". With the Cartesian convention the focal length of a concave mirror is negative, and that one slip turns a real image into a virtual one.

Pause and predict: you hold a concave shaving mirror close to your face, between the pole and the focus. Will the image be erect or inverted, real or virtual?

Check your answer: erect and virtual, and magnified. With $u$ between $0$ and $f$, the mirror equation gives a positive $v$, so the image is behind the mirror, and $m = -v/u$ is positive. That is why a shaving mirror magnifies only when you stay close.

Beyond the textbook. For the exam: a spherical mirror sends all parallel rays to the focus. In reality: that holds only for paraxial rays. Rays far from the axis cross the axis closer to the mirror and blur the image, an effect called spherical aberration. Headlamps and large telescope mirrors are made parabolic, which focuses all parallel rays to one point. Why it matters: the formulas in this chapter are first order approximations for small apertures.

Refraction: light changes speed, so it bends

When light crosses from one medium to another, it changes speed, and a slanted ray changes direction. Snell's law states this:

$$n_1 \sin i = n_2 \sin r, \qquad \text{or} \qquad \dfrac{\sin i}{\sin r} = n_{21} = \dfrac{n_2}{n_1}$$

The refractive index of a medium is $n = c/v$, the ratio of the speed of light in vacuum to its speed in the medium. Going into a medium with a larger n (a slower medium), the ray bends towards the normal. Going into a smaller n, it bends away. For two media $n_{12} = 1/n_{21}$, and for three media $n_{32} = n_{31} \times n_{12}$.

Optical density is not mass density. A medium is optically denser if its refractive index is larger, meaning light travels more slowly in it. Turpentine has less mass per volume than water but a higher refractive index, so it is optically denser.

For a parallel sided slab, the emerging ray is parallel to the incoming ray. It is shifted sideways but not turned. For an object under a transparent medium viewed nearly head on:

$$\text{apparent depth} = \dfrac{\text{real depth}}{n}$$

So a 2.0 m deep pool ($n = 1.33$) looks only $2.0/1.33 = 1.5 \text{ m}$ deep, which answers our opening question.

Total internal reflection

Suppose light goes from a denser medium to a rarer one. The ray bends away from the normal. Increase the angle of incidence, and the refracted ray finally skims along the surface at 90°. The angle of incidence at which this happens is the critical angle $i_c$:

$$\sin i_c = \dfrac{n_{\text{rarer}}}{n_{\text{denser}}} = \dfrac{1}{n} \qquad (\text{for a rarer medium of air})$$

For any larger angle there is no refracted ray at all, and the light is completely reflected. Two conditions are needed: the light must be travelling from denser to rarer, and the angle of incidence must exceed $i_c$. Total internal reflection never happens when light enters a denser medium.

TIR is used in right angled prisms (bending light by 90° or by 180°), in optical fibres and in endoscopes. A fibre has a core of higher refractive index than its cladding. Light entering at a suitable angle strikes the core–cladding boundary beyond the critical angle again and again and is carried along the fibre, even round bends, with very little loss.

Worked example: a bulb under water. A small bulb sits at the bottom of a tank 80 cm deep, filled with water of refractive index $4/3$. Find the area of the water surface through which light can escape into the air.

Common wrong step. Writing $r = h \sin i_c$. The radius is the horizontal distance the ray covers in a right triangle where the vertical side is h, so it is $h \tan i_c$.

Pause and predict: can light going from air to glass ever be totally internally reflected?

Check your answer: no. TIR needs light to travel from a denser medium to a rarer one. Entering glass from air, the ray always bends towards the normal, and a refracted ray always exists, even for an incidence of 90°.

Beyond the textbook. For the exam: a diamond sparkles because its critical angle (24.4°) is small, so most light inside it is totally reflected. In reality: the small critical angle is only part of the story. Jewellers cut a diamond with carefully chosen facet angles so that light entering the top is reflected twice inside and exits through the top. Diamond also separates colours strongly, which gives the "fire". Why it matters: the critical angle explains why TIR is easy in diamond, and the cut explains why you see it.

Refraction at a spherical surface and thin lenses

For a single spherical surface of radius R between media of index $n_1$ (object side) and $n_2$, paraxial rays obey

$$\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2 - n_1}{R}$$

A thin lens is two such surfaces back to back. Refracting at the first surface and using its image as the object for the second gives the lens maker's formula:

$$\dfrac{1}{f} = (n - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$$

where $n$ is the index of the lens material relative to its surroundings and $R_1$, $R_2$ are the radii of the first and second surfaces. For a double convex lens $R_1$ is positive and $R_2$ is negative, so $f$ is positive. For a double concave lens $f$ is negative. The thin lens formula and magnification are

$$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}, \qquad m = \dfrac{v}{u}$$

Compare this carefully with the mirror equation, which has a plus sign in $1/v + 1/u$. Lens: minus. Mirror: plus.

A concave lens always forms a virtual, erect, diminished image between the lens and its focus.

Power of a lens. The power measures how strongly a lens bends light:

$$P = \dfrac{1}{f} \quad \text{with } f \text{ in metres}, \qquad \text{unit dioptre (D)}, \qquad 1 \text{ D} = 1 \text{ m}^{-1}$$

A converging lens has positive power and a diverging one has negative power. For thin lenses in contact:

$$\dfrac{1}{f} = \dfrac{1}{f_1} + \dfrac{1}{f_2} + \cdots, \qquad P = P_1 + P_2 + \cdots, \qquad m = m_1 \times m_2 \times \cdots$$

This works because the first lens's image acts as the object for the second, and for thin lenses in contact the distances are measured from effectively the same point.

Worked example: image by a convex lens. An object is 30 cm in front of a convex lens of focal length 20 cm. Find the image distance and magnification.

The image is real, inverted, twice the size, and 60 cm on the far side of the lens.

Common wrong step. Using the mirror equation's sign, $1/v + 1/u = 1/f$, which gives $1/v = 1/20 + 1/30$ and $v = 12 \text{ cm}$. Always remember that the lens equation has a minus sign.

Worked example: lens maker and combination. A double convex lens made of glass of refractive index 1.5 has radii of curvature 10 cm and 15 cm. (a) Find its focal length. (b) It is placed in contact with a concave lens of focal length 20 cm. Find the focal length and power of the combination.

Common wrong step. Putting the focal lengths into $P = 1/f$ in centimetres. That gives answers 100 times too small, so convert to metres before using dioptres.

Pause and predict: a glass lens ($n = 1.5$) is dipped in water ($n = 1.33$). Does its focal length increase or decrease?

Check your answer: it increases. In the lens maker's formula, n becomes the ratio of glass to water, about 1.13, and $(n - 1)$ falls from 0.5 to about 0.13. The lens bends light far less, and f becomes roughly four times longer.

Pause and predict: the lower half of a convex lens is covered with an opaque card. What happens to the image?

Check your answer: the whole image still forms, but it is dimmer. Every point of the object sends rays to every part of the lens, and the uncovered half is enough to form the full image. Only the amount of light gathered is halved.

Beyond the textbook. For the exam: a thin lens has one fixed focal length. In reality: the refractive index depends on colour, so a simple lens focuses blue light a little nearer than red. This is chromatic aberration, and the lens maker's formula gives f for one colour only. Camera and microscope lenses use several elements of different glasses to cancel it, which is also why mirror telescopes have an advantage. Why it matters: it explains why instruments are far more complicated than the single lens in a diagram.

Refraction through a prism

A prism has two refracting faces separated by the angle of the prism $A$. A ray enters at incidence angle $i$, refracts to $r_1$, meets the second face at $r_2$, and leaves at angle $e$. Geometry gives

$$r_1 + r_2 = A, \qquad d = i + e - A$$

As $i$ increases from a small value, the deviation $d$ first decreases, reaches a minimum $D_m$, and then increases. At the minimum the ray passes symmetrically through the prism: $i = e$ and $r_1 = r_2 = A/2$, so the ray inside is parallel to the base. Applying Snell's law at the first face:

$$n = \dfrac{\sin[(A + D_m)/2]}{\sin(A/2)}$$

For a thin prism, with a small A, the deviation is

$$\delta = (n - 1)A$$

Worked example: prism at minimum deviation. An equilateral prism has a refractive index of $\sqrt{2}$. Find the angle of minimum deviation and the angle of incidence at that setting.

Common wrong step. Using the thin prism formula $\delta = (n - 1)A$, which gives 24.8°. That formula is valid only for very small prism angles, not for an equilateral prism.

Optical instruments

Simple microscope (magnifying glass). A single convex lens of short focal length, used with the object just inside the focus. The comfortable viewing distance is $D = 25 \text{ cm}$, the near point of a normal eye.

$$m = 1 + \dfrac{D}{f} \quad (\text{image at the near point}), \qquad m = \dfrac{D}{f} \quad (\text{image at infinity})$$

The second form is the more relaxing for the eye, because the final image is at infinity. The two differ by exactly 1.

Compound microscope. Two lenses do the work. The objective, a lens of very short focal length $f_0$, makes a real, inverted and magnified image of a tiny object. The eyepiece, with focal length $f_e$, then acts as a simple microscope on that image. The total magnification is the product of the two:

$$m = \left(\dfrac{L}{f_0}\right)\left(1 + \dfrac{D}{f_e}\right) \quad (\text{final image at the near point}), \qquad m = \left(\dfrac{L}{f_0}\right)\left(\dfrac{D}{f_e}\right) \quad (\text{final image at infinity})$$

Here $L$ is the distance between the objective's second focal point and the eyepiece's first focal point, the tube length. Both focal lengths are small and $L$ is large, which gives big magnification.

Refracting telescope. The objective has a large focal length $f_0$ and a large aperture. It collects light from a far object and forms a real image at its focus. A short focal length eyepiece then magnifies that image. In normal adjustment (final image at infinity):

$$m = \dfrac{f_0}{f_e}, \qquad \text{tube length} = f_0 + f_e$$

Terrestrial telescopes add an erecting lens, because the image of a simple astronomical telescope is inverted.

Reflecting (Cassegrain) telescope. A large concave primary mirror collects the light, and a small convex secondary mirror sends it back through a hole in the primary to the eyepiece. Mirrors have no chromatic aberration, can be supported from behind so they can be very large, are lighter than lenses of the same size, and can be made parabolic to remove spherical aberration. The Keck telescopes in Hawaii have 10 m mirrors, and the largest in India is the 2.34 m telescope at Kavalur.

Worked example: instruments. (a) A compound microscope has an objective of focal length 1.0 cm, an eyepiece of focal length 2.0 cm, and a tube length of 20 cm. Find the magnification with the final image at infinity and at the near point ($D = 25 \text{ cm}$). (b) A telescope has $f_0 = 100 \text{ cm}$ and $f_e = 4 \text{ cm}$. Find its magnifying power, its length, and the size of the image of the Moon, whose angular diameter is 0.5°.

Common wrong step. Taking the telescope magnification as $f_e/f_0$, which gives 0.04 and suggests the instrument shrinks the Moon. The long focal length lens is the objective, and it goes on top.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Electromagnetic Waves shows that visible light is an EM wave of tiny wavelength, which is why the ray picture works.
• Next: Wave Optics shows what happens when the ray picture fails: interference, diffraction and polarisation.
• Class XI link: Waves gives the idea of wave speed that explains why light bends when its speed changes.
• Later link: Dual Nature of Radiation and Matter uses the particle picture of the same light.