Ray Optics — Concepts, Formulas & Solved Numericals

Complete guide to ray optics for Class 12. Solved examples, exam tips, PYQs.

CBSE-12JEE-MAINNEET14 min readClasses 12

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 vv in a medium, then n=c/vn = c/v where c=3×108c = 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/fP = 1/f in dioptres (D), where ff 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=ri = 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.

Snell's Law

n1sini=n2sinrn_1 \sin i = n_2 \sin r

Or equivalently: sinisinr=n2n1=1n2\dfrac{\sin i}{\sin r} = \dfrac{n_2}{n_1} = {}_1n_2

The relative refractive index 1n2{}_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.

Critical Angle Formula

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

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

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

💡 Expert Tip

Diamonds are cut at angles specifically so that light undergoes TIR multiple times inside before exiting. The high refractive index of diamond (n2.42n \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.

Mirror Formula

1v+1u=1f=2R\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.

Magnification (Mirror)

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

Positive mm → virtual, erect image. Negative mm → real, inverted image.

Refraction at Spherical Surfaces

Refraction at Spherical Surface

n2vn1u=n2n1R\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

Thin Lens Formula

1v1u=1f\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}

Lensmaker's Equation

1f=(n1)(1R11R2)\dfrac{1}{f} = (n-1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)

R1R_1 = radius of the first surface, R2R_2 = radius of the second surface. Use the Cartesian sign convention carefully here.

Power of a Lens System

P=P1+P2+P3+P = P_1 + P_2 + P_3 + \ldots

For lenses in contact, powers add directly.

Refraction Through a Prism

Prism Formula

n=sin ⁣(A+D2)sin ⁣(A2)n = \dfrac{\sin\!\left(\dfrac{A+D}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)}

where AA = prism angle, DD = angle of minimum deviation.

At minimum deviation, the ray passes symmetrically through the prism and r1=r2=A/2r_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=20u = -20 cm, f=15f = -15 cm

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

1v=1f1u=115120=115+120\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15} - \frac{1}{-20} = -\frac{1}{15} + \frac{1}{20}

1v=4+360=160\frac{1}{v} = \frac{-4 + 3}{60} = \frac{-1}{60}

v=60 cmv = -60 \text{ cm}

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

m=vu=6020=3m = -\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=1.5n = 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°i = 90° - 60° = 30°.

Applying Snell's law (air to glass):

1×sin30°=1.5×sinr1 \times \sin 30° = 1.5 \times \sin r

sinr=0.51.5=13\sin r = \frac{0.5}{1.5} = \frac{1}{3}

r=sin1 ⁣(13)19.47°r = \sin^{-1}\!\left(\frac{1}{3}\right) \approx 19.47°

⚠️ Common Mistake

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 (nw=4/3n_w = 4/3).

(a) In air:

For a biconvex lens, R1=+20R_1 = +20 cm and R2=30R_2 = -30 cm (applying sign convention).

1f=(n1)(1R11R2)=(1.51)(120130)\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×(120+130)=0.5×3+260=0.5×560=124= 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 cmf = 24 \text{ cm}

(b) In water:

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

1fw=(1.1251)(120+130)=0.125×560=0.62560\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}

fw=600.625=96 cmf_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

🎯 Exam Insider

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.

🎯 Exam Insider

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 dd: use 1f=1f1+1f2df1f2\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}.

🎯 Exam Insider

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.


Common Mistakes to Avoid

⚠️ Common Mistake

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

⚠️ Common Mistake

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|m|. A magnification of 3-3 means 3× magnified, inverted.

⚠️ Common Mistake

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.

⚠️ Common Mistake

Mistake 4: Confusing power addition with focal length addition. Powers add directly for lenses in contact. Focal lengths do NOT add. If f1=20f_1 = 20 cm and f2=30f_2 = 30 cm, the combined focal length is NOT 50 cm. Calculate P1+P2P_1 + P_2 and take the reciprocal.

⚠️ Common Mistake

Mistake 5: Using prism formula at the wrong condition. The formula n=sin ⁣(A+D2)/sin ⁣(A2)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?


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


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


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?


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


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


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?


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?


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, ff is positive and uu is always negative. Substituting in the mirror formula always gives vv positive (image behind mirror) and v<u|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 (n1)(n - 1) where nn is the refractive index of glass relative to the surrounding medium. In water, the effective nn drops from 1.5 (in air) to about 1.125, making (n1)(n-1) smaller and ff 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|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 nn values in problems sometimes specify the colour of light.

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

Apparent depth =real depthn= \frac{\text{real depth}}{n}. For water (n=1.33n = 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+DfM = 1 + \frac{D}{f} when the image is at the near point (D=25D = 25 cm), or M=D/fM = 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.

Practice Questions

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