Ray Optics — Concepts, Formulas & Solved Numericals

Ray Optics — The Geometry of Light

Light travels in straight lines. That single assumption is the foundation of ray optics, and it’s surprisingly powerful. We use it to explain why mirrors form images, why a straw looks bent in water, and how a lens can focus sunlight enough to start a fire.

Ray optics (also called geometric optics) treats light as rays — directed lines showing the path of energy. We ignore the wave nature of light entirely here. That simplification lets us use clean geometry and algebra to solve problems that would otherwise require heavy physics.

This chapter has a weightage of 8-10 marks in CBSE board exams and 3-4 questions in JEE Main. NEET asks 2-3 questions, mostly from refraction, lenses, and the human eye. Every mark here is predictable — the question types repeat year after year.

Key Terms and Definitions

Incident ray — the ray hitting a surface. Reflected ray — the ray bouncing back. Refracted ray — the ray that enters the second medium (and bends, usually).

Normal — an imaginary line perpendicular to the surface at the point of incidence. All angles in optics are measured from the normal, not the surface. This trips up students constantly.

Angle of incidence (i) — angle between incident ray and normal.

Angle of reflection (r) — angle between reflected ray and normal.

Angle of refraction (r’) — angle between refracted ray and normal inside the second medium.

Refractive index (n or μ) — how much slower light travels in a medium compared to vacuum. If light travels at speed $v$ in a medium, then $n = c/v$ where $c = 3 \times 10^8$ m/s.

Critical angle (C) — the angle of incidence (in a denser medium) at which refracted ray just grazes the surface (r’ = 90°).

Optical centre (P) — the geometric centre of a lens.

Focal length (f) — distance from optical centre to the focus.

Power of a lens (P) — $P = 1/f$ in dioptres (D), where $f$ is in metres.

Core Laws and Concepts

Laws of Reflection

Both laws hold for any surface — plane, convex, or concave.

Law 1: Incident ray, reflected ray, and normal all lie in the same plane.

Law 2: Angle of incidence = angle of reflection. $i = r$ .

For a plane mirror, the image is:

Virtual and erect
Same size as the object
As far behind the mirror as the object is in front

Refraction and Snell’s Law

When light goes from one medium to another, it bends. Why? Because speed changes at the boundary. The ray bends toward the normal when entering a denser medium (slower speed), and away when entering a rarer medium.

n_1 \sin i = n_2 \sin r

Or equivalently: $\dfrac{\sin i}{\sin r} = \dfrac{n_2}{n_1} = {}_1n_2$

The relative refractive index ${}_1n_2$ tells us how dense medium 2 is relative to medium 1.

Total Internal Reflection (TIR)

TIR happens only when light travels from denser to rarer medium AND the angle of incidence exceeds the critical angle.

\sin C = \dfrac{1}{n}

where $n$ is the refractive index of the denser medium with respect to air.

Applications of TIR: optical fibres, diamond sparkle, mirage, periscopes using prisms.

Diamonds are cut at angles specifically so that light undergoes TIR multiple times inside before exiting. The high refractive index of diamond ( $n \approx 2.42$ ) gives a very small critical angle (~24°), making this easy to achieve.

Spherical Mirrors

Concave mirror — reflecting surface curves inward (cave-like). Converging.

Convex mirror — reflecting surface curves outward. Diverging.

\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} = \dfrac{2}{R}

Sign convention: distances measured from the pole (P). Distances in the direction of incident light are positive; opposite are negative.

m = -\dfrac{v}{u} = \dfrac{h_i}{h_o}

Positive $m$ → virtual, erect image. Negative $m$ → real, inverted image.

Refraction at Spherical Surfaces

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

This formula is used for a single refracting surface — a lens is just two such surfaces combined.

Thin Lens Formula and Lensmaker’s Equation

\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}

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

$R_1$ = radius of the first surface, $R_2$ = radius of the second surface. Use the Cartesian sign convention carefully here.

P = P_1 + P_2 + P_3 + \ldots

For lenses in contact, powers add directly.

Refraction Through a Prism

n = \dfrac{\sin\!\left(\dfrac{A+D}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)}

where $A$ = prism angle, $D$ = angle of minimum deviation.

At minimum deviation, the ray passes symmetrically through the prism and $r_1 = r_2 = A/2$ .

Solved Examples

Example 1 — Easy (CBSE Level)

An object is placed 20 cm in front of a concave mirror of focal length 15 cm. Find the image position and magnification.

Using mirror formula with sign convention (mirror at origin, object to the left):

$u = -20$ cm, $f = -15$ cm

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15} - \frac{1}{-20} = -\frac{1}{15} + \frac{1}{20}

\frac{1}{v} = \frac{-4 + 3}{60} = \frac{-1}{60}

v = -60 \text{ cm}

Image is 60 cm in front of the mirror (real, since $v$ is negative).

m = -\frac{v}{u} = -\frac{-60}{-20} = -3

Image is 3× magnified and inverted.

Example 2 — Medium (JEE Main Level)

A ray of light falls on a glass slab ( $n = 1.5$ ) at an angle of 60° with the surface. Find the angle of refraction.

The angle with the surface is 60°, so the angle of incidence with the normal is $i = 90° - 60° = 30°$ .

Applying Snell’s law (air to glass):

1 \times \sin 30° = 1.5 \times \sin r

\sin r = \frac{0.5}{1.5} = \frac{1}{3}

r = \sin^{-1}\!\left(\frac{1}{3}\right) \approx 19.47°

Many students directly use 60° as the angle of incidence. The angle is always measured from the normal. If the problem gives the angle with the surface, subtract from 90° first.

Example 3 — Hard (JEE Advanced Level)

A biconvex lens has radii of curvature 20 cm and 30 cm. The refractive index of glass is 1.5. Find: (a) focal length in air, (b) focal length when immersed in water ( $n_w = 4/3$ ).

(a) In air:

For a biconvex lens, $R_1 = +20$ cm and $R_2 = -30$ cm (applying sign convention).

\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (1.5 - 1)\left(\frac{1}{20} - \frac{1}{-30}\right)

= 0.5 \times \left(\frac{1}{20} + \frac{1}{30}\right) = 0.5 \times \frac{3 + 2}{60} = 0.5 \times \frac{5}{60} = \frac{1}{24}

f = 24 \text{ cm}

(b) In water:

The effective refractive index is now $n_{\text{eff}} = n_{\text{glass}} / n_{\text{water}} = 1.5 / (4/3) = 1.5 \times 3/4 = 1.125$ .

\frac{1}{f_w} = (1.125 - 1)\left(\frac{1}{20} + \frac{1}{30}\right) = 0.125 \times \frac{5}{60} = \frac{0.625}{60}

f_w = \frac{60}{0.625} = 96 \text{ cm}

The lens is still converging but much weaker in water. This is why fish can see without goggles — their eye lens is adapted for a water-glass interface, not air-glass.

Exam-Specific Tips

CBSE Board Pattern: Ray optics is in Chapter 9, Class 12. Expect one 3-mark numerical (mirror/lens formula), one 2-mark conceptual (TIR or image characteristics), and sometimes a diagram-based question. Marking scheme is step-marking — show the formula, substitution, and final answer clearly.

JEE Main: Questions frequently mix two concepts — a common combo is “refraction at spherical surface + lens formula” or “TIR + optical fibre”. JEE Main 2024 Shift 1 had a question on the equivalent focal length of two thin lenses separated by a distance $d$ : use $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$ .

NEET: Stick to conceptual clarity. NEET asks about the human eye (accommodation, defects, corrective lenses), microscope/telescope magnification, and simple mirror/lens numericals. The prism formula appears occasionally but TIR and optical fibre applications are almost guaranteed.

For SAT Physics Subject Test: Focus on the qualitative behaviour — what happens to image size and position as object moves. Memorise the sign convention for virtual vs real images.

Real-World Examples

Example 1: Your Spectacles Are a Thin Lens Formula Problem

A Class 12 student in Pune visits an optometrist and walks out with a prescription: −2.5 D for the left eye. That negative sign tells the whole story — she is myopic (short-sighted), and her eye’s lens is converging light too strongly, forming the image in front of the retina. The spectacle lens prescribed is a concave lens of focal length $f = \frac{1}{P} = \frac{1}{-2.5} = -0.4\ \text{m}$ , which diverges incoming rays just enough to shift the image back onto the retina.

Connect to the syllabus: This is a direct application of the lens-maker’s power formula $P = \frac{1}{f}$ and the correction-of-defects-of-vision section — a favourite 2-mark theory question in CBSE 12 board exams.

Example 2: The DSLR Camera Lens and Real Inverted Images

When a photographer at an IPL match in the Wankhede Stadium uses a 300 mm telephoto lens, the lens forms a tiny, real, inverted image of the batsman on the camera sensor. The object distance $u$ is roughly 50 m; the image distance $v$ works out to just over 300 mm — consistent with $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (using the sign convention). The sensor then digitally flips and magnifies it. Every selfie and news photograph you see is an inverted real image first.

Connect to the syllabus: This is the $u > 2f$ case for a convex lens where the image is real, inverted, and diminished — exactly the ray diagram that appears in JEE Main and NEET almost every year.

Example 3: The Periscope on INS Kalvari

India’s submarine INS Kalvari, operating in the Arabian Sea, uses a periscope to observe surface ships while staying submerged. Inside the periscope, two plane mirrors are fixed at exactly $45°$ to the vertical, separated by about 10 m. Light from a surface target hits the top mirror, reflects downward, hits the bottom mirror, and exits horizontally into the observer’s eye — with the angle of incidence equal to the angle of reflection at each surface. No lens is needed for the basic design; pure law of reflection does the job.

Connect to the syllabus: This is a direct, exam-ready application of the law of reflection ( $\angle i = \angle r$ ) and the geometry of successive reflections from inclined plane mirrors — a classic short-answer in CBSE and an occasional NEET MCQ context question.

Common Mistakes to Avoid

Mistake 1: Using focal length directly without sign convention. For a concave mirror, $f$ is negative. For a convex lens, $f$ is positive. For a diverging (concave) lens, $f$ is negative. Always assign signs before substituting.

Mistake 2: Forgetting that magnification can be negative. Negative magnification doesn’t mean the image is smaller. It means the image is inverted. Size is $|m|$ . A magnification of $-3$ means 3× magnified, inverted.

Mistake 3: Applying TIR to the wrong direction. TIR only occurs when going from denser to rarer medium. Light going from water to air can undergo TIR. Light going from air to water cannot, no matter the angle.

Mistake 4: Confusing power addition with focal length addition. Powers add directly for lenses in contact. Focal lengths do NOT add. If $f_1 = 20$ cm and $f_2 = 30$ cm, the combined focal length is NOT 50 cm. Calculate $P_1 + P_2$ and take the reciprocal.

Mistake 5: Using prism formula at the wrong condition. The formula $n = \sin\!\left(\frac{A+D}{2}\right) / \sin\!\left(\frac{A}{2}\right)$ is only valid at minimum deviation. If the problem doesn’t specify minimum deviation, use Snell’s law at each surface separately.

Practice Questions

Q1. A convex mirror has a focal length of 20 cm. Where is an object placed if its image is formed 10 cm behind the mirror?

Using mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

For convex mirror: $f = +20$ cm, $v = +10$ cm (image behind mirror).

$\frac{1}{u} = \frac{1}{f} - \frac{1}{v} = \frac{1}{20} - \frac{1}{10} = \frac{1-2}{20} = \frac{-1}{20}$

$u = -20$ cm. Object is 20 cm in front of the mirror.

Q2. What is the critical angle for glass with $n = 1.732$ ?

$\sin C = \frac{1}{n} = \frac{1}{1.732} = \frac{1}{\sqrt{3}}$

$C = 30°$

Q3. A ray enters a prism of angle 60° at one face and exits at the other face. If the refractive index is $\sqrt{3}$ , find the angle of minimum deviation.

Using $n = \frac{\sin\!\left(\frac{A+D}{2}\right)}{\sin\!\left(\frac{A}{2}\right)}$

$\sqrt{3} = \frac{\sin\!\left(\frac{60°+D}{2}\right)}{\sin 30°} = \frac{\sin\!\left(\frac{60°+D}{2}\right)}{0.5}$

$\sin\!\left(\frac{60°+D}{2}\right) = \frac{\sqrt{3}}{2}$

$\frac{60+D}{2} = 60°$ , so $D = 60°$ .

Q4. A concave lens of focal length 30 cm is placed in contact with a convex lens of focal length 20 cm. What is the power and nature of the combination?

$P_1 = \frac{1}{-0.30} = -3.33$ D (concave lens, negative)

$P_2 = \frac{1}{0.20} = +5$ D

$P = P_1 + P_2 = -3.33 + 5 = +1.67$ D

The combination is converging (convex equivalent) with $f = 1/1.67 \approx 60$ cm.

Q5. An object 5 cm tall is placed 25 cm from a convex lens of focal length 10 cm. Find the image height.

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ , $u = -25$ cm, $f = +10$ cm

$\frac{1}{v} = \frac{1}{10} + \frac{1}{-25} = \frac{1}{10} - \frac{1}{25} = \frac{5-2}{50} = \frac{3}{50}$

$v = \frac{50}{3} \approx 16.7$ cm

$m = \frac{v}{u} = \frac{50/3}{-25} = -\frac{2}{3}$

Image height $= m \times h_o = -\frac{2}{3} \times 5 = -3.33$ cm (real, inverted, smaller).

Q6. Light travels from glass ( $n = 1.5$ ) to water ( $n = 1.33$ ). What is the critical angle?

At critical angle, refracted ray goes along the surface (r = 90°).

$n_{\text{glass}} \sin C = n_{\text{water}} \sin 90°$

$\sin C = \frac{n_{\text{water}}}{n_{\text{glass}}} = \frac{1.33}{1.5} = 0.887$

$C = \sin^{-1}(0.887) \approx 62.5°$

Q7. A person uses a convex lens of power +2 D as a reading glass. What is the focal length? Where should a book be placed to get a clear image at 25 cm?

$f = \frac{1}{P} = \frac{1}{2} = 0.5$ m $= 50$ cm.

We want the virtual image at 25 cm, so $v = -25$ cm (same side as object for virtual image from convex lens).

$\frac{1}{u} = \frac{1}{v} - \frac{1}{f}$ … wait, use lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{-25} - \frac{1}{50} = \frac{-2-1}{50} = \frac{-3}{50}$

$u = -\frac{50}{3} \approx -16.7$ cm.

The book should be placed about 16.7 cm from the lens.

Q8. A glass slab of thickness 6 cm and refractive index 1.5 is placed over a coin. By how much does the coin appear shifted upward?

Apparent shift $= t\left(1 - \frac{1}{n}\right) = 6\left(1 - \frac{1}{1.5}\right) = 6 \times \frac{1}{3} = 2$ cm.

The coin appears 2 cm closer (shifted up by 2 cm).

FAQs

What is the difference between real and virtual images?

A real image is formed when reflected/refracted rays actually converge at a point. You can project it on a screen. A virtual image is formed where the rays appear to diverge from — the rays don’t actually meet there. Plane mirrors, convex mirrors, and concave lenses always form virtual images. Concave mirrors and convex lenses form real images when the object is beyond the focal point.

Why does a convex mirror always form a virtual, erect, and diminished image?

For a convex mirror, $f$ is positive and $u$ is always negative. Substituting in the mirror formula always gives $v$ positive (image behind mirror) and $|v| < |u|$ , so the image is always virtual, erect, and smaller than the object. This is why convex mirrors are used as rear-view mirrors — wider field of view with the diminished image.

What happens to the focal length of a lens when it’s immersed in water?

The focal length increases. Why? The lensmaker’s equation uses $(n - 1)$ where $n$ is the refractive index of glass relative to the surrounding medium. In water, the effective $n$ drops from 1.5 (in air) to about 1.125, making $(n-1)$ smaller and $f$ larger. The lens becomes weaker.

Can a concave mirror form a virtual image?

Yes, when the object is placed between the pole and the focus (i.e., $|u| < |f|$ ). In this case, the reflected rays diverge and appear to come from a point behind the mirror. This is how a concave mirror works as a shaving mirror — the face is placed within the focal length.

What is dispersion and is it in ray optics?

Dispersion — the splitting of white light into a spectrum by a prism — is technically in this chapter. The refractive index of glass is different for different wavelengths (violet bends most, red least). This is why $n$ values in problems sometimes specify the colour of light.

Why is the bottom of a pool always shallower than its actual depth?

Apparent depth $= \frac{\text{real depth}}{n}$ . For water ( $n = 1.33$ ), a 4-metre pool looks only 3 metres deep. The refraction of light at the water-air interface makes objects beneath the surface appear closer.

What is the magnifying power of a simple microscope?

$M = 1 + \frac{D}{f}$ when the image is at the near point ( $D = 25$ cm), or $M = D/f$ when the image is at infinity. For NEET, you must know both formulas and when to use each — if the problem says “image at least distance of distinct vision”, use the first formula.

How do optical fibres use TIR?

An optical fibre has a dense glass core surrounded by a rarer cladding. Light entering at one end hits the core-cladding boundary at angles greater than the critical angle, so it undergoes TIR repeatedly along the entire length of the fibre — no energy loss at each bounce. This is how data travels as light pulses over thousands of kilometres.