Ray Optics and Optical Instruments

The chapter builds outward from two experimental laws toward working optical instruments. It opens with reflection by spherical mirrors: the Cartesian sign convention, the proof that focal length is half the radius of curvature, and the mirror equation with magnification. It turns next to refraction — Snell's law, lateral shift through a slab, apparent depth — before a full section on total internal reflection, the critical angle, and optical fibres. The middle derives refraction at a single spherical surface and uses it twice to build the lens maker's formula and thin lens equation, followed by power and lenses in contact. A section on the prism derives the deviation formula and the minimum-deviation route to refractive index. The chapter closes with optical instruments — simple magnifier, compound microscope, refracting and reflecting (Cassegrain) telescopes — all read through angular magnification. Four worked examples, extending NCERT's own numbers, and a final cheat sheet consolidate the chapter for revision.

Reflection at spherical mirrors and the Cartesian sign convention

The law of reflection is familiar from Class 10: the angle of incidence equals the angle of reflection, and the incident ray, reflected ray and normal all lie in the same plane. This holds at every point of any reflecting surface, plane or curved — for a curved mirror the normal at a point is simply the line joining that point to the mirror's centre of curvature $C$, since the normal is always perpendicular to the tangent to the surface. The geometric centre of the mirror is its pole $P$, and the line through $P$ and $C$ is the principal axis. To turn geometry into algebra, this chapter fixes one Cartesian sign convention. All distances are measured from the pole of the mirror (or optical centre of a lens) along the principal axis: measured along the direction of incident light they are positive, against it negative. Heights above the axis are positive, below are negative. This single convention lets one mirror equation and one lens equation handle concave, convex, real and virtual images alike, with no separate formula for each case. A parallel beam striking a concave mirror converges, after reflection, to a point $F$ on the axis, the principal focus; a convex mirror makes the reflected rays appear to diverge from an $F$ behind the mirror. The distance $PF$ is the focal length $f$, and the first derivation below proves it is exactly half the radius of curvature $R$.

The mirror equation and magnification

To locate an image it is enough to trace two easy rays from a point on the object: one parallel to the axis, reflecting through the focus; one through the centre of curvature, retracing its own path. Doing this for every object point and applying the sign convention leads to the mirror equation,

where: $u$ is the object distance, $v$ is the image distance, and $f$ is the focal length, all measured from the pole along the sign convention above. Because $f$ and $u$ are negative for a real object at a mirror, Equation (1) automatically gives the correct sign for $v$ — negative for a real image in front of the mirror, positive for a virtual image behind it (proved fully in the second derivation below). Alongside it, the linear magnification

where: $h$ and $h'$ are the object and image heights (positive upward, negative downward by the sign convention), tells you both the size and the orientation of the image in one number: a negative $m$ means an inverted, real image, while a positive $m$ means an erect, virtual image.

Refraction and Snell's law

When light crosses from one transparent medium into another at an angle, it bends at the interface — refraction. Snell's experimental laws state that the incident ray, refracted ray and normal are coplanar, and that the ratio of sines is constant for a given pair of media:

where: $i$ is the angle of incidence, $r$ is the angle of refraction, both from the normal, and $n_{21}$ is the refractive index of medium 2 relative to medium 1. If $n_{21} > 1$ the ray bends toward the normal ($r < i$) and medium 2 is optically denser; if $n_{21} < 1$ it bends away and medium 2 is rarer. Optical density is not mass density — turpentine is optically denser than water despite lower mass density, since optical density is the ratio of the speed of light in the two media. Reversing the pair inverts the index, $n_{12} = 1/n_{21}$, and chaining three media gives $n_{32} = n_{31} \times n_{12}$. Two consequences follow immediately. A ray crossing a parallel-sided glass slab twice emerges parallel to its original direction but laterally displaced, since the two refractions cancel the angular bending. And a submerged object viewed from directly above appears raised: for near-normal viewing, apparent depth equals real depth divided by the refractive index of the medium, which is why a pool always looks shallower than it is.

Total internal reflection, critical angle and optical fibres

Refraction at a denser-to-rarer interface bends the ray away from the normal, so refraction grows faster than incidence. At one particular angle, the critical angle $i_c$, the angle of refraction reaches exactly $90^\circ$ and the refracted ray grazes the interface. Push $i$ any further and refraction becomes impossible — every bit of light reflects back into the denser medium: total internal reflection, which unlike ordinary reflection loses no intensity to transmission at all. Applying Snell's law at $r = 90^\circ$ gives

where: $n_{21}$ is the refractive index of the rarer medium relative to the denser one ($n_{21} < 1$), and $n_{12} = 1/\sin i_c$. Typical critical angles with respect to air: about $48.75^\circ$ for water ($n=1.33$), $41.14^\circ$ for crown glass ($n=1.52$), $37.31^\circ$ for dense flint glass ($n=1.62$), and only $24.41^\circ$ for diamond ($n=2.42$) — its small critical angle is why a well-cut diamond traps and bounces light so effectively. Total internal reflection redirects light without loss. A right-angled prism with critical angle below $45^\circ$ (true for both crown and dense flint glass) bends a beam by $90^\circ$ or $180^\circ$, or inverts an image without changing size, using total internal reflection at its hypotenuse — cleaner than a silvered mirror. The key application is the optical fibre: a core of high refractive index glass surrounded by lower-index cladding. Light entering suitably strikes the core-cladding boundary beyond the critical angle every time, reflecting totally along the entire length even around bends, emerging with almost no loss. Modern silica fibres transmit over $95\%$ of light across a full kilometre, which is why fibres, not copper wire, carry the world's telephone and internet traffic, and serve as light pipes for medical endoscopes.

Refraction at a spherical surface and thin lenses

The same reasoning used for a spherical mirror, applied instead to a single spherical refracting surface separating two media of refractive index $n_1$ and $n_2$, gives an analogous relation between object and image distance:

where: $u$ and $v$ are the object and image distances from the surface, $R$ its radius of curvature, and $n_1$, $n_2$ the refractive indices of the medium light leaves and enters. Applying Equation (5) twice, once at each face of a thin lens (worked in full in the third derivation below), produces the lens maker's formula,

where: $f$ is the focal length, $n_2/n_1$ the refractive index of the lens material relative to its surroundings, and $R_1$, $R_2$ the signed radii of the two surfaces (positive and negative respectively for a symmetric double convex lens). The same derivation gives the everyday thin lens equation,

which looks like the mirror equation but has a different sign pattern, since refraction does not reverse the direction of light. It holds for convex and concave lenses, real and virtual images alike, with magnification

Three rays trace an image without algebra: one parallel to the axis refracts through the second focal point; one through the optical centre passes undeviated; one through the first focal point emerges parallel to the axis.

Power of a lens and combinations of lenses in contact

A lens of short focal length bends light sharply; one of long focal length bends it gently. This is captured as the lens's power,

where: $f$ is in metres and $P$ in dioptre (D), $1\ \text{D} = 1\ \text{m}^{-1}$, positive for converging and negative for diverging lenses — a $+2.5\ \text{D}$ prescription is a convex lens of $f=+40\ \text{cm}$, a $-4.0\ \text{D}$ one a concave lens of $f=-25\ \text{cm}$. When several thin lenses of focal lengths $f_1, f_2, \ldots$ share an optical centre in contact, the image of one becomes the object for the next, and adding the individual thin lens equations gives an equivalent single lens

Powers add algebraically, so a convex and concave lens in contact partially cancel, converging only if the net power stays positive; the total magnification is the product $m = m_1 m_2 \cdots$. Such stacks, chosen to add power while cancelling each other's defects, are the backbone of camera lenses and microscope objectives.

Refraction through a prism

A triangular prism refracts at two faces. Light enters face $AB$ at $i$, refracts to $r_1$ inside the glass, strikes face $AC$ from inside at $r_2$, and emerges at $e$. The net bending between original and final direction is the angle of deviation $\delta$. Prism geometry (derived in full below) gives

where: $A$ is the prism angle. As $i$ varies, $\delta$ falls to a single minimum $D_m$ and rises again; by the symmetry of Equation (11) under swapping $i$ and $e$, this minimum occurs at $i=e$, forcing $r_1=r_2=A/2$. At this minimum deviation, the internal ray runs parallel to the base, and Snell's law at the first face gives the prism formula

turning two measurable angles into the refractive index — the standard spectrometer method. For a thin prism, small $A$ and $D_m$ collapse Equation (12) to $D_m \approx (n_{21} - 1)A$: deviation proportional to prism angle.

Optical instruments — the simple and compound microscope

A simple microscope, or magnifying glass, is a single converging lens of short focal length held close to both eye and object, so it sits just inside the focal length and forms an erect, magnified, virtual image. The closest comfortable viewing distance is the near point, $D \approx 25\ \text{cm}$. With the image at the near point,

while relaxing the eye so the image forms at infinity gives the slightly smaller but more comfortable

where: $f$ is the lens focal length. A simple magnifier does not enlarge the object; it lets the object be brought closer than $25\ \text{cm}$ so it subtends a larger angle. Realistic focal lengths cap useful magnification around $9\times$. For higher magnification, a compound microscope uses two lenses. The objective, nearest the object, forms a real, inverted, magnified image; that image, lying near the focal plane of the eyepiece, is then viewed exactly as a simple microscope views an object. Writing $m_o$ for the objective's linear magnification, $m_e$ for the eyepiece's angular magnification, and $L$ for the tube length (distance between the objective's second focal point and the eyepiece's first), the approximate formula for a final image at infinity is

where: $f_o$, $f_e$ are the objective and eyepiece focal lengths, both kept short — which is why real objectives and eyepieces are compact multi-element lens systems rather than single thin lenses.

Optical instruments — the telescope

A telescope magnifies the angle a distant object subtends, using an objective of large focal length and aperture to form a real image, which the eyepiece then magnifies as a simple microscope would:

where: $f_o$, $f_e$ are the objective and eyepiece focal lengths, and the tube length in normal adjustment is $f_o + f_e$. Light-gathering and resolving power both improve with objective diameter, but a large refracting lens is heavy, hard to support, and costly to make free of chromatic aberration — so every large modern telescope is a reflecting telescope, using a mirror objective instead: no chromatic aberration, less weight, full-back support. The Cassegrain design uses a convex secondary mirror to reflect light back through a hole in the primary, giving a long effective focal length in a short tube. India's largest, a $2.34\ \text{m}$ Cassegrain at Kavalur, is run by the Indian Institute of Astrophysics; the world's largest, the Keck telescopes in Hawaii, use $10\ \text{m}$ mirrors.

Board level - locating the image in a concave mirror (extends NCERT Example 9.3)

A candle is placed 10 cm in front of a concave mirror of radius of curvature 15 cm. Find the position, nature and magnification of the image, and the height of the image if the candle flame is 1.2 cm tall.

Step 1. Find the focal length: $f = R/2 = 15/2 = 7.5\ \text{cm}$; by the sign convention for a concave mirror, $f = -7.5\ \text{cm}$. The object distance is also negative, $u = -10\ \text{cm}$.

Step 2. Apply the mirror equation: $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{-7.5} - \dfrac{1}{-10} = -0.1333 + 0.1 = -0.0333\ \text{cm}^{-1}$, giving $v = -30\ \text{cm}$.

Step 3. Find the magnification: $m = -v/u = -(-30)/(-10) = -3$. The negative sign of $v$ means the image forms $30\ \text{cm}$ in front of the mirror, on the same side as the object, so it is real; the negative $m$ means it is inverted, and $|m| = 3$ means it is three times the size of the object.

Step 4. Find the image height: $h' = m \times h = -3 \times 1.2 = -3.6\ \text{cm}$. Answer: the image forms 30 cm in front of the mirror, is real, inverted and magnified three times, with height 3.6 cm.

Board level - lens maker's formula and power (extends NCERT Example 9.7)

A double convex lens has radii of curvature 10 cm and 15 cm, and its focal length is measured to be 12 cm. Find the refractive index of the glass, and the power of the lens.

Step 1. Assign signed radii for a double convex lens with the first surface bulging toward the incoming light and the second bulging away from it: $R_1 = +10\ \text{cm}$, $R_2 = -15\ \text{cm}$.

Step 2. Compute the geometric factor: $\dfrac{1}{R_1} - \dfrac{1}{R_2} = \dfrac{1}{10} - \left(\dfrac{1}{-15}\right) = \dfrac{1}{10} + \dfrac{1}{15} = \dfrac{3+2}{30} = \dfrac{1}{6}\ \text{cm}^{-1}$.

Step 3. Substitute into the lens maker's formula with $f = +12\ \text{cm}$: $\dfrac{1}{12} = (n-1)\times\dfrac{1}{6}$, so $n - 1 = \dfrac{6}{12} = 0.5$, giving $n = 1.5$.

Step 4. Convert the focal length to metres to find the power: $f = 0.12\ \text{m}$, so $P = 1/f = 1/0.12 = 8.33\ \text{D}$. Answer: the refractive index of the glass is 1.5, and the power of the lens is +8.33 dioptre (a converging lens).

Competitive level (JEE / NEET) - prism at minimum deviation combined with total internal reflection

A glass prism has a refracting angle $A = 60^\circ$ and shows minimum deviation $D_m = 30^\circ$ for a monochromatic ray. Find (a) the refractive index of the glass, (b) the critical angle for the glass-air interface, and (c) what happens to a ray that enters the first face normally and then strikes the second face.

Step 1. Apply the prism formula: $n = \dfrac{\sin[(A+D_m)/2]}{\sin(A/2)} = \dfrac{\sin 45^\circ}{\sin 30^\circ} = \dfrac{0.7071}{0.5} = 1.414 = \sqrt{2}$.

Step 2. Find the critical angle from $\sin i_c = 1/n$: $\sin i_c = 1/\sqrt{2} = 0.7071$, so $i_c = 45^\circ$.

Step 3. Trace the specific ray. Normal incidence at the first face means $i = 0$, so by Snell's law $r_1 = 0$ as well — the ray travels straight through undeviated to the second face. By the prism geometry $r_1 + r_2 = A$, so $r_2 = A - r_1 = 60^\circ - 0^\circ = 60^\circ$.

Step 4. Compare $r_2$ with the critical angle found in Step 2: since $r_2 = 60^\circ$ is greater than $i_c = 45^\circ$, the ray striking the second face from inside the denser prism at $60^\circ$ cannot refract out at all — it undergoes total internal reflection at the second face instead of emerging. Answer: $n = \sqrt{2} \approx 1.414$, the critical angle is $45^\circ$, and a ray entering normally at the first face is totally internally reflected at the second face rather than emerging through it. Timing note: under two minutes. The trap is stopping after part (b) — always check whether the actual angle of incidence at the second face (found from simple geometry, not from another minimum-deviation calculation) exceeds the critical angle before assuming the ray emerges; this exact combination, prism geometry feeding into a total-internal-reflection check, is a favourite two-in-one JEE question.

Competitive level (JEE / NEET) - compound microscope with final image at infinity (extends NCERT Exercise 9.11)

A compound microscope has an objective of focal length $f_o = 2.0\ \text{cm}$ and an eyepiece of focal length $f_e = 6.25\ \text{cm}$, with the two lenses separated by $15\ \text{cm}$. If the final image is formed at infinity, find the distance of the object from the objective and the magnifying power of the microscope. Take $D = 25\ \text{cm}$.

Step 1. For the final image to form at infinity, the intermediate real image formed by the objective must sit exactly at the eyepiece's first focal point, a distance $f_e$ in front of the eyepiece. Since the two lenses are $15\ \text{cm}$ apart, this intermediate image forms at $v_1 = 15 - 6.25 = 8.75\ \text{cm}$ from the objective.

Step 2. Apply the thin lens equation to the objective to find the object distance $u_1$: $\dfrac{1}{v_1} - \dfrac{1}{u_1} = \dfrac{1}{f_o}$, so $\dfrac{1}{u_1} = \dfrac{1}{v_1} - \dfrac{1}{f_o} = \dfrac{1}{8.75} - \dfrac{1}{2} = 0.1143 - 0.5 = -0.3857\ \text{cm}^{-1}$, giving $u_1 = -2.59\ \text{cm}$.

Step 3. Find the magnification produced by the objective alone: $m_o = v_1/u_1 = 8.75/(-2.59) = -3.375$.

Step 4. Find the eyepiece's angular magnification for an image at infinity, $m_e = D/f_e = 25/6.25 = 4$, and combine: $m = m_o \times m_e = -3.375 \times 4 = -13.5$. Answer: the object must be placed 2.59 cm in front of the objective, and the magnifying power of the microscope is 13.5, with the final image inverted relative to the object. Timing note: about two to three minutes. The trap is treating the given 15 cm lens separation as the "tube length $L$" in the approximate formula $m \approx (L/f_o)(D/f_e)$ — that formula's $L$ is the distance between the objective's second focal point and the eyepiece's first focal point, not the physical lens separation, and using the wrong one gives a magnification badly off from the exact lens-equation answer computed here. When both numbers are given, always solve exactly with the thin lens equation rather than reaching for the approximate formula.

• Applying the mirror sign convention to lenses, or vice versa. A real image has $v$ negative for a mirror but positive for a lens. Mixing the two conventions is the single most common source of wrong signs in this chapter.
• Forgetting that $f = R/2$ carries its own sign. A concave mirror has both $R$ and $f$ negative; a convex mirror has both positive.
• Confusing $n_{21}$ with $n_{12}$. $n_{21} = \sin i/\sin r$ is the index of the medium entered relative to the medium left; swap the subscripts and every critical-angle or prism calculation inverts.
• Assuming total internal reflection can happen going from a rarer to a denser medium. It cannot — only denser-to-rarer travel bends the refracted ray away from the normal fast enough to disappear entirely.
• Forgetting $R_2$ is measured from the second surface, not the first. For a double convex lens, $R_1$ (along the incident light) is positive while $R_2$ (against it) is negative, even though both centres of curvature are real points.
• Mixing up the tube length $L$ in the compound microscope formula with the physical lens separation. They coincide only in idealised diagrams; when a real separation is given, solve exactly with the thin lens equation, as in the fourth worked example.

• State the Cartesian sign convention precisely and apply it consistently to both mirrors and lenses without confusing the two.
• Derive the relation $f = R/2$ for a spherical mirror from paraxial-ray geometry.
• Use the mirror equation and magnification formula to find the position, nature and size of an image formed by a concave or convex mirror.
• State Snell's law, explain lateral shift through a slab and apparent depth, and compute a critical angle from a refractive index.
• Explain, with a labelled diagram, how total internal reflection makes an optical fibre work, and why it requires denser-to-rarer travel above the critical angle.
• Derive the lens maker's formula from refraction at a single spherical surface, and use the thin lens equation and power for single lenses and combinations in contact.
• Derive the prism formula $n_{21} = \sin[(A+D_m)/2]/\sin(A/2)$ and use it to find a refractive index from measured angles.
• Distinguish linear magnification from angular magnification (magnifying power) for a simple microscope, and compute both.
• Calculate the magnifying power of a compound microscope and of a refracting or reflecting telescope, given the focal lengths and, where needed, the exact lens separation.
• Explain why large modern telescopes are reflecting (Cassegrain) telescopes rather than refracting ones.

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