Ray Optics and Optical Instruments

Take a prism ABC with refracting angle $A$ at vertex A, between faces AB and AC. A ray PQ strikes face AB at Q with angle of incidence $i$ and refracts at angle $r_1$; it crosses the glass to R on face AC, meets that face from inside at angle $r_2$, and emerges at the angle of emergence $e$. Let the normals drawn at Q and R meet at N.

First, a fact about the angles inside. In the quadrilateral AQNR, the angles at Q and R are both 90° (each normal is perpendicular to its face), and the four angles of a quadrilateral add up to 360°, so the remaining two must add up to 180°:

In triangle QNR, the interior angles at Q and R are exactly $r_1$ and $r_2$ (each is the angle between the ray inside the glass and a normal), so:

Comparing the two equations:

Now the deviation. At the first face the ray is turned through $i - r_1$; at the second it is turned through $e - r_2$, in the same sense. Adding them and using $r_1 + r_2 = A$:

Here $e$ is not a free choice: once $i$ is set, Snell’s law at the first face fixes $r_1$ ($\sin i = n\sin r_1$), $r_1 + r_2 = A$ fixes $r_2$, and Snell’s law at the second face fixes $e$ ($\sin e = n\sin r_2$). So for a given prism, $\delta$ is a function of $i$ alone. Plotted against $i$, it is largest at the two extremes (grazing incidence, or grazing emergence) and dips to a single minimum value, the angle of minimum deviation $\delta_m$, in between.

Where exactly is that minimum? Light paths are reversible: send a ray backwards along the emergent path and it retraces the whole route, now entering at angle $e$ and leaving at angle $i$, and since $\delta = i + e - A$ is symmetric in $i$ and $e$, with exactly the same deviation. So every deviation larger than the minimum is produced by two different angles of incidence, $i$ and $e$. At the very bottom of the curve there is only one point, so those two values must merge into one (sometimes written $i_1 = i_2$):

With $i = e$, Snell’s law at the two faces ($\sin i = n\sin r_1$, $\sin e = n\sin r_2$) forces $r_1 = r_2 = r$ as well: the ray passes through the prism symmetrically, and for the usual isosceles prism it travels parallel to the base. Now the two geometric results of §9.9.1 pin everything down. From $r_1 + r_2 = A$:

and from $\delta = i + e - A$ with $\delta = \delta_m$:

Substituting both into Snell’s law at the first face, $n = \sin i/\sin r$:

Both $A$ and $\delta_m$ can be measured directly on a spectrometer, which makes this the standard laboratory method for measuring the refractive index of a glass. Strictly, the $n$ here is $n_{21}$, the prism’s index relative to whatever surrounds it; that matters as soon as the prism is put in a liquid (NCERT Exercise 9.6, below).

For a thin prism (refracting angle of only a few degrees) with light arriving close to the normal, every angle involved is small, so Snell’s law at each face becomes $i \approx n r_1$ and $e \approx n r_2$. Then:

independent of the angle of incidence, to this approximation: a thin prism turns every near-normal ray through the same small angle.

The refractive index of glass is not one fixed number: it depends on the wavelength of the light. Refractive index is $n = c/v$, and the speed of light inside glass depends on its frequency (the glass’s bound electrons respond more strongly to higher-frequency light, slowing it more), so in ordinary glass violet light travels more slowly than red, and $n_v > n_r$. In a vacuum every colour travels at the same speed $c$, so a vacuum shows no such effect. Since a prism’s deviation depends on $n$, each colour in white light is deviated by a different amount: red least, violet most. White light entering a prism therefore comes out fanned into a spectrum of colours. This splitting of light into its constituent colours is dispersion.

For a thin prism, each colour obeys §9.9.2’s $\delta = (n - 1)A$ with its own refractive index:

The angle between the emerging violet and red rays, the angular dispersion, is the difference:

It grows with the prism angle and with how strongly the glass’s refractive index varies across the spectrum.

The cornea and the eye lens together act as a converging lens system that forms a real, inverted image on the retina. Unlike a camera, though, the eye cannot move its lens closer to or further from its “screen”: the lens-to-retina distance is fixed. To focus on objects at different distances, the eye changes its focal length instead. The ciliary muscles change the curvature of the flexible eye lens, which is exactly the knob §9.5’s lens maker’s formula says controls $f$. This ability is called accommodation. With the ciliary muscles fully relaxed the lens is at its flattest and weakest, and the eye is focused on its far point, which for a normal eye is at infinity. At maximum accommodation the lens bulges to its most curved and strongest, and the eye is focused on its near point, the closest point it can see clearly. For a normal adult eye this is taken as $D = 25\text{ cm}$, the least distance of distinct vision, the same $D$ that appears in every magnifying-power formula of §9.11.

How much accommodation does that take? Model the eye as a thin lens a fixed distance $v_r$ in front of the retina, so the image distance never changes. Focused on infinity ($u \to -\infty$) and on the near point ($u = -D$), the thin-lens formula gives the eye’s power in each state:

whatever the actual size of the eye: a normal eye must be able to add about 4 dioptres to its own power. Every defect of vision below is a far point, a near point, or an accommodation range that fails to match this.

A myopic eye sees nearby objects clearly but distant ones blurred: its far point lies at some finite distance $x$ instead of at infinity. Either the eyeball is too long or the relaxed eye lens is too powerful, so parallel rays from a distant object converge to a focus in front of the retina. The eye has too much converging power, so the fix must take some away: a diverging (concave) lens.

Exactly which lens? It must take parallel rays from an object at infinity and make them diverge as if they came from the far point, the furthest point this eye can focus on. Treating the spectacle lens as being in contact with the eye, that means $u = -\infty$ and a virtual image at $v = -x$:

The corrective lens’s focal length is simply minus the far-point distance, negative, as the argument above demanded.

A hypermetropic eye sees distant objects clearly but its near point has receded to some distance $y$ beyond 25 cm, so a book held at a normal reading distance blurs. Either the eyeball is too short or the eye lens, even at full accommodation, is too weak, so rays from a nearby object would converge to a focus behind the retina. The eye has too little converging power, so the fix must add some: a converging (convex) lens.

The lens must take an object held at the normal near point $D$ and form a virtual image of it at the defective near point $y$, where this eye can focus. So $u = -D$ and $v = -y$:

Since $y > D$, $f$ is positive: converging, as argued. With $D = 0.25\text{ m}$ the power is $P = 4 - 1/y$ dioptres, which correctly falls to zero (no correction needed) when $y = D$.

With age, the eye lens gradually stiffens and the ciliary muscles weaken, so the eye slowly loses its power of accommodation: the roughly 4 D range of §9.10.1 shrinks, and the near point recedes, typically becoming noticeable somewhere in a person’s forties. This is presbyopia. Its effect on near vision is the same as hypermetropia, but its cause is different: not an eyeball of the wrong length, but a lost range of focusing. The correction for reading is therefore the same converging lens, with power given by §9.10.3’s formula. Presbyopia can also coexist with myopia, in which case the same person needs a diverging lens to see far and a converging lens to read. The standard answer is a bifocal lens: a diverging upper half for distance and a converging lower half for reading, where the eyes naturally look down.

A single convex lens used close to the eye, forming a magnified, virtual, erect image. With the image formed at the near point $D$ (usually 25 cm), its magnifying power is:

With the image formed at infinity (relaxed-eye viewing, the more comfortable setting), magnifying power is simply $m=D/f$.

A simple magnifier’s power, $D/f$, grows as its focal length shrinks, but very short focal-length lenses are hard to grind and badly distort the image, so a single lens tops out at a magnification of about 9 for realistic focal lengths. To go further, a compound microscope uses two converging lenses, each multiplying the effect of the other: an objective of short focal length $f_o$ close to the object, and an eyepiece of short focal length $f_e$ close to the eye.

The object is placed just beyond the objective’s focal point, so the objective forms a real, inverted, magnified intermediate image, of height $h'$, inside the tube. That intermediate image is positioned at (or just inside) the eyepiece’s first focal point, so the eyepiece works exactly like the simple microscope of §9.11.1 with the intermediate image as its object, producing an enlarged, virtual final image either at infinity or at the near point. The final image is inverted relative to the original object. The two lenses magnify in genuinely different ways, so each needs its own treatment.

The objective: a linear magnification. It forms a real image, so what matters is actual size: by §9.6’s sign convention its linear magnification is $m_o = h'/h = v_o/u_o$. To bring the focal length in, multiply the thin-lens formula through by $v_o$:

Here $v_o - f_o$ is the distance from the objective’s second focal point to the intermediate image. The tube length $L$ is defined as the distance between the objective’s second focal point and the eyepiece’s first focal point, and since the intermediate image sits at (or very near) that eyepiece focal point, $v_o - f_o \approx L$:

exact when the final image is at infinity, and the minus sign in $m_o$ simply records that the intermediate image is inverted.

The eyepiece: an angular magnification. The eyepiece makes a virtual image, so what matters is the angle the final image subtends at the eye, not its size. Its magnifying power is the simple-microscope result of §9.11.1, applied to the intermediate image:

Combining them. The magnifying power of the whole microscope compares the angle $\beta$ subtended at the eye by the final image with the angle the object itself would subtend if held at the near point and viewed unaided, $\theta_o = h/D$. The eyepiece, by definition of its own magnifying power, makes the final image subtend $m_e$ times the angle the intermediate image would subtend at $D$, so $\beta = m_e\,(h'/D)$. Therefore:

The objective’s linear magnification and the eyepiece’s angular magnification simply multiply. For the two standard adjustments:

Both formulas reward a short objective focal length, a short eyepiece focal length and a long tube, which is why both lenses of a microscope have short focal lengths (in practice it is hard to make them much shorter than about 1 cm). When a numerical problem gives the actual separation of the lenses, it is more accurate to find $|m_o| = v_o/|u_o|$ from the lens formula directly than to use $L/f_o$, as the worked example below shows, and as NCERT’s own answers do.

Two converging lenses, an objective of large focal length $f_o$ and an eyepiece of small focal length $f_e$, used to magnify distant objects. In normal adjustment (final image at infinity), magnifying power is:

A longer objective and a shorter eyepiece both increase magnifying power, which is why telescope objectives are built as large (long focal length) as practically possible.