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Optics: JEE Main Physics formulas and notes

Optics is one of the biggest chapters in JEE Mains Physics: ray optics (mirrors, refraction, TIR, lenses, prism, instruments) and wave optics (Huygens, YDSE, diffraction, polarisation) together give 3 to 4 questions every year. The formulas are many but standard; most marks are lost to sign errors, using the wrong focal length in a liquid, or confusing the fringe width with the position of a fringe. About 20 hours is enough.

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Formulas with solved examples

Reflection and mirrors

Plane mirror

How to apply. Draw the image behind the mirror at equal distance. For rotation use 2θ. For two mirrors find 360/θ first, then subtract 1 when it is even.

Example. Mirrors at 60°: n = 360 ÷ 60 − 1 = 5 images. A mirror turned by 15° turns the reflected ray by 30°. A person walks towards a fixed mirror at 2 m/s: the image approaches him at 4 m/s.

Spherical mirror formula and sign convention

How to apply. Write u (always negative for a real object), f with its sign (negative for concave, positive for convex), then find 1/v = 1/f − 1/u. Use m = −v/u to find size and nature.

Example. Concave mirror f = −20 cm, object at u = −30 cm: 1/v = −1/20 + 1/30 = −1/60, v = −60 cm (real); m = −(−60)/(−30) = −2: inverted, twice as large. Convex mirror f = +15 cm, u = −30 cm: 1/v = 1/15 + 1/30, v = +10 cm (virtual), m = +1/3.

Magnification and the moving object

How to apply. Find m at the given position, then speed of image = m² × speed of object, in the same direction as the object’s speed (for mirrors the image moves the same way along x).

Example. Concave f = −20 cm, object at 30 cm so m = −2; object moves at 2 cm/s towards the mirror: image speed = 4 × 2 = 8 cm/s (away from the mirror, since v = −60 cm grows in size).

Refraction and total internal reflection

Snell’s law and speed of light in a medium

How to apply. Measure both angles from the normal. Apply n₁ sin i = n₂ sin r, then v = c/n and λ = λ₀/n if speed or wavelength is asked.

Example. Light from air at 60° enters glass n = 1.5: sin r = 0.866 ÷ 1.5 = 0.577, r = 35.3°. In glass v = 2 × 10⁸ m/s and a 600 nm wave has λ = 400 nm.

Apparent depth and slab

How to apply. Divide each real thickness by its index (for an observer in air) and add. The shift is t(1 − 1/n), directed towards the observer.

Example. A fish 24 cm below the surface of water (n = 4/3) appears at 18 cm. A coin under a 6 cm glass slab (n = 1.5) looks raised by 6(1 − 2/3) = 2 cm.

Lateral shift in a glass slab

How to apply. Find r from Snell’s law, then compute sin(i − r) and cos r.

Example. Slab t = 3 cm, n = 1.5, i = 45°: sin r = 0.7071 ÷ 1.5 = 0.471, r = 28.1°; d = 3 × sin(16.9°) ÷ cos 28.1° = 3 × 0.290 ÷ 0.882 = 0.99 cm.

Total internal reflection and critical angle

How to apply. Check that the ray goes from a higher to a lower index. Compare the incident angle with θc. For a source below a surface, the light that escapes fills a circle of radius R.

Example. Water n = 4/3: sinθc = 0.75, θc = 48.6°. A lamp 7 cm deep gives a bright circle of radius R = 7 ÷ √(16/9 − 1) = 7.94 cm. Glass n = 1.5: θc = 41.8°.

Optical fibre

How to apply. Compute n₁² − n₂² and take the square root; the inverse sine is the largest angle of entry measured from the axis.

Example. n₁ = 1.5, n₂ = 1.4: NA = √(2.25 − 1.96) = 0.539, θ(max) = 32.6°.

Refraction at a spherical surface

How to apply. Fix signs: u negative, R positive for a surface convex towards the object. Substitute into n₂/v − n₁/u = (n₂ − n₁)/R and solve for v.

Example. Air to glass (n = 1.5), R = +15 cm, object at u = −45 cm: 1.5/v = 0.5/15 + 1/(−45) = 1/30 − 1/45 = 1/90, v = +135 cm (real image). For u = −30 cm the image is at infinity.

Lenses

Thin lens formula and lens maker’s formula

How to apply. Take the first surface the light meets as R₁. Use signs: R positive when the centre is on the outgoing side. Get f from the maker’s formula, then use 1/v = 1/f + 1/u.

Example. Biconvex lens R₁ = +20 cm, R₂ = −20 cm, n = 1.5: 1/f = 0.5 × (1/20 + 1/20) = 0.05, f = 20 cm, P = +5 D. In water (4/3): n = 1.125 and 1/f = 0.125 × 0.1, f = 80 cm. For u = −30 cm in air: 1/v = 1/20 − 1/30, v = 60 cm, m = −2.

Image positions and the minimum distance

How to apply. Place the object relative to f and 2f to name the image. For object–screen distance D > 4f there are two lens positions.

Example. f = 20 cm and the object at 40 cm gives a real inverted image of the same size at 40 cm, 80 cm = 4f from the object. An object at 15 cm gives 1/v = 1/20 − 1/15, v = −60 cm: virtual, m = +4.

Combinations of lenses

How to apply. Add powers directly if the lenses touch. Otherwise take the image of the first lens as the object of the second, or use the d formula for the equivalent focal length.

Example. +10 cm and −20 cm in contact: 1/f = 0.1 − 0.05 = 0.05, f = +20 cm, P = 10 − 5 = +5 D. Two lenses of f = 20 cm separated by 10 cm: 1/f = 1/20 + 1/20 − 10/400 = 0.075, f = 13.3 cm.

Cut lenses, silvered lens and displacement method

How to apply. Decide whether the lens is cut along or across the axis. For a silvered lens add the powers: twice for the lens and once for the mirror surface.

Example. Displacement method: D = 90 cm, d = 30 cm: f = (8100 − 900) ÷ 360 = 20 cm. A symmetric biconvex lens of f = 20 cm cut across its thickness gives two plano-convex lenses of f = 40 cm each.

Prism and dispersion

Prism: deviation and refractive index

How to apply. Draw the ray; use r₁ + r₂ = A inside. At minimum deviation take i = (A + δₘ)/2. Check whether the ray emerges (r₂ < θc).

Example. A = 60°, δₘ = 30°: n = sin 45° ÷ sin 30° = 1.414. Thin prism A = 4°, n = 1.5: δ = 2°. Normal incidence, A = 30°, n = √2: sin e = 0.707, e = 45°, δ = 15°.

Prism: limits of emergence

How to apply. Compute θc = sin⁻¹(1/n) and compare with the angle r₂ (or A, for normal incidence).

Example. n = 1.5 (θc = 41.8°): a ray can emerge from a prism only if A ≤ 83.6°; for normal incidence the ray emerges from the second face only if A < 41.8°.

Dispersion and dispersive power

How to apply. For a single prism find δ for each colour and subtract. For an achromatic pair equate the angular dispersions of the two prisms; the deviation left over is (n − 1)A − (n′ − 1)A′.

Example. nᵣ = 1.51, nᵥ = 1.53, A = 5°: angular dispersion = 0.02 × 5 = 0.1°; n = 1.52, ω = 0.02 ÷ 0.52 = 0.038. Crown (A = 10°, nᵥ − nᵣ = 0.016, n = 1.52) with flint (nᵥ′ − nᵣ′ = 0.032, n′ = 1.65): A′ = 5° removes dispersion; net deviation 5.2° − 3.25° = 1.95°.

Scattering and natural phenomena

How to apply. Compare wavelengths and raise the ratio to the fourth power. For a bow, the observer’s back is towards the Sun.

Example. Violet (400 nm) versus red (700 nm): (700/400)⁴ = 9.4, so violet is scattered about 9.4 times more.

Optical instruments and the eye

Simple microscope (magnifier)

How to apply. Put the image where asked: near point gives 1 + D/f, infinity gives D/f. D = 25 cm unless given otherwise.

Example. f = 5 cm: M = 1 + 25/5 = 6 at the near point and 5 at infinity.

Compound microscope

How to apply. Find v₀ from the lens formula for the objective, m₀ = v₀/u₀, then Mₑ for the final position. Multiply.

Example. f₀ = 1 cm, fₑ = 5 cm, object at 1.1 cm: 1/v₀ = 1 − 1/1.1, v₀ = 11 cm, m₀ = −10. Image at infinity: M = 10 × 25/5 = 50 (tube length 11 + 5 = 16 cm). Image at 25 cm: M = 10 × 6 = 60.

Astronomical telescope

How to apply. Use L = f₀ + fₑ for normal adjustment. For near-point viewing find the eyepiece distance uₑ from the lens formula with v = −D.

Example. f₀ = 100 cm, fₑ = 5 cm: M = −20 and L = 105 cm. Final image at 25 cm: M = −20 × 1.2 = −24, uₑ = 125 ÷ 30 = 4.17 cm, L = 104.2 cm.

Terrestrial, Galilean and reflecting telescopes

How to apply. For the Galilean telescope subtract the focal lengths: the eyepiece is a concave lens placed inside the focus of the objective.

Example. Galilean telescope with f₀ = 30 cm and fₑ = 3 cm: M = 10 and L = 27 cm.

Resolving power

How to apply. Apply the right formula to the instrument: use the diameter D₀ for a telescope and the numerical aperture for a microscope. Keep λ in metre.

Example. A telescope with a 5 m mirror at λ = 550 nm: θ = 1.22 × 550 × 10⁻⁹ ÷ 5 = 1.34 × 10⁻⁷ rad. Microscope with n sinβ = 0.8 and λ = 600 nm: d = 1.22 × 600 × 10⁻⁹ ÷ 1.6 = 4.6 × 10⁻⁷ m. Eye of pupil 2 mm, λ = 500 nm: θ = 3.05 × 10⁻⁴ rad.

Defects of vision

How to apply. Place the object where it should be seen (infinity for myopia, 25 cm for hypermetropia) and the image at the actual far or near point. Then use the lens formula 1/f = 1/v − 1/u.

Example. Far point 80 cm: f = −80 cm, P = −1.25 D. Near point 50 cm: f = 25 × 50 ÷ 25 = 50 cm, P = +2 D (check: 1/v − 1/u = −1/50 + 1/25 = 1/50).

Wavefronts and interference

Huygens’ principle, wavefronts and optical path

How to apply. Convert every path to optical path (n x) and then to phase using φ = 2π Δ/λ. A glass plate of thickness t adds (n − 1)t to the path.

Example. A glass sheet n = 1.5 and t = 3 μm in a path adds (0.5)(3 μm) = 1.5 μm. For λ = 600 nm that is 2.5 wavelengths, a phase difference 5π.

Superposition and intensity of two waves

How to apply. Convert path difference to phase: φ = 2πΔ/λ. Use the general formula for unequal sources or 4I₀cos²(φ/2) for equal ones. For a ratio of I use a = √(I₁/I₂).

Example. I₁ : I₂ = 9 : 1: a = 3, I(max)/I(min) = (4/2)² = 4. Two equal sources I₀ with φ = 60°: I = 4I₀cos²30° = 3I₀.

Young’s double slit: fringes

How to apply. Write Δ = dy/D and set it equal to nλ (bright) or (n + ½)λ (dark). Fringe width is the distance between two neighbouring bright fringes.

Example. λ = 600 nm, D = 1.5 m, d = 0.3 mm: β = 6 × 10⁻⁷ × 1.5 ÷ 3 × 10⁻⁴ = 3 mm. The 4th bright fringe is at 12 mm. Under water (4/3) β = 2.25 mm.

Shift of the fringes by a thin sheet

How to apply. Find the extra optical path (n − 1)t, divide by λ to get the number of fringes N, and multiply by β to get the shift.

Example. Sheet n = 1.5, t = 6 μm, λ = 600 nm, β = 3 mm: N = 0.5 × 6 × 10⁻⁶ ÷ 6 × 10⁻⁷ = 5 fringes, shift = 5 × 3 = 15 mm toward the covered slit.

Two wavelengths and white light

How to apply. Find the smallest integers n₁ : n₂ = λ₂ : λ₁. The distance of the first coincidence is n₁β₁.

Example. λ₁ = 600 nm, λ₂ = 500 nm: n₁ : n₂ = 5 : 6; the first coincidence is at the 5th fringe of 600 nm (the 6th of 500 nm), y = 5 × 3 mm = 15 mm.

Diffraction and resolution

Single slit diffraction

How to apply. Set a sinθ = λ for the first minimum, with y = D tanθ ≈ Dθ. The central maximum extends from −y₁ to +y₁.

Example. a = 0.2 mm, λ = 500 nm, D = 1 m: first minimum at y = λD/a = 2.5 mm; width of the central maximum = 5 mm.

Double slit with finite slit width

How to apply. Find d/a; the order that equals this ratio is missing. The central envelope holds all orders below it on each side, plus the central one.

Example. d = 0.6 mm and a = 0.1 mm (d/a = 6): the 6th order is missing, so the central diffraction maximum has 2 × 6 − 1 = 11 bright fringes.

Polarisation

Polarisation and Malus’ law

How to apply. The first polariser always halves unpolarised light. After that multiply by cos² of the angle between successive axes.

Example. Unpolarised I₀ through polarisers at 0°, 30° and 90°: I = (I₀/2)cos²30°cos²60° = (I₀/2)(3/4)(1/4) = 3I₀/32. With only two crossed polarisers the light is zero; a third at 45° between them gives I₀/8.

Brewster’s law

How to apply. Check whether the angle is Brewster’s; use tan i = n or the fact that the reflected and refracted rays are perpendicular.

Example. Glass n = 1.5: iʙ = tan⁻¹1.5 = 56.3° and r = 33.7°. Water n = 4/3: iʙ = 53.1°.

Thin films

Thin-film interference and coatings

How to apply. Count the λ/2 phase jumps: one jump (air to film) means reflected maxima are at 2nt = (m + ½)λ. For a coating on glass both reflections jump, so the minimum for darkness is 2nt = λ/2.

Example. Soap film n = 1.33, bright in reflected light for λ = 560 nm with the smallest thickness: t = 560 ÷ (4 × 1.33) = 105 nm. A MgF₂ coating (n = 1.38) for 550 nm on glass: t = 550 ÷ (4 × 1.38) = 99.6 nm.

Most asked question types

How do I solve “Plane mirrors: rotation, images and speed” questions in JEE Main?

A mirror turns, two mirrors meet at an angle or a person moves in front of a mirror. Find the angle, the number of images, or the speed of the image. 1. Mirror turned by θ: the reflected ray turns by 2θ. 2. Images between mirrors at θ: 360/θ − 1 (even case). 3. The image is as far behind as the object is in front. Example: Mirrors at 60°: 5 images.

How do I solve “Spherical mirror: image position and magnification” questions in JEE Main?

An object is placed in front of a concave or convex mirror. Find the image distance, the magnification, the nature of the image or the focal length. 1. Write u and f with their signs. 2. 1/v = 1/f − 1/u; m = −v/u. 3. Sign of v and m decide real or virtual, erect or inverted. Example: Concave f = −20 cm, u = −30 cm: v = −60 cm, m = −2.

How do I solve “Refraction, apparent depth and slab shift” questions in JEE Main?

Light passes into a medium or an object is viewed through water or glass. Find the angle, the apparent depth, the shift or the speed. 1. Use Snell’s law with the angles from the normal. 2. Apparent depth = real depth/n. 3. Shift = t(1 − 1/n). Example: Fish 24 cm deep in water appears at 18 cm.

How do I solve “Total internal reflection and critical angle” questions in JEE Main?

Light goes from a denser to a rarer medium, or a source is placed under a liquid. Find the critical angle, whether TIR occurs, or the radius of the light circle. 1. sinθc = n₂/n₁. 2. Compare the angle of incidence with θc. 3. R = h/√(n² − 1) for a source at depth h. Example: Water, source 7 cm deep: R = 7.94 cm.

How do I solve “Lens formula and lens maker’s formula” questions in JEE Main?

A lens or a combination of surfaces is given with radii and index. Find f, the image distance, the magnification, the power, or the effect of immersion in a liquid. 1. 1/f = (n − 1)(1/R₁ − 1/R₂) with signs. 2. 1/v = 1/f + 1/u; m = v/u. 3. In a liquid replace n by n(lens)/n(liquid). Example: Biconvex R = 20 cm, n = 1.5: f = 20 cm; in water 80 cm.

How do I solve “Combinations: lenses in contact, separated and mirrors” questions in JEE Main?

Two lenses touch or are placed some distance apart, or a lens is silvered. Find the net focal length, the power or the final image. 1. In contact: P = P₁ + P₂. 2. Separated: image of the first is the object of the second. 3. Silvered lens: P = 2P(lens) + P(mirror). Example: +10 cm and −20 cm in contact: f = 20 cm, P = 5 D.

How do I solve “Prism: deviation, minimum deviation and refractive index” questions in JEE Main?

A ray passes through a prism. Find the angle of deviation, the angle of emergence, the refractive index from δₘ, or the condition for emergence. 1. δ = i + e − A and r₁ + r₂ = A. 2. At minimum deviation n = sin((A + δₘ)/2)/sin(A/2). 3. Check whether r₂ is less than θc. Example: A = 60°, δₘ = 30°: n = 1.414.

How do I solve “Dispersion and dispersive power” questions in JEE Main?

The indices for red and violet are given with the angle of a thin prism. Find the angular dispersion, the dispersive power, or the angle of the second prism of an achromatic pair. 1. θ = (nᵥ − nᵣ)A. 2. ω = (nᵥ − nᵣ)/(n − 1). 3. Achromatic: (nᵥ − nᵣ)A = (nᵥ′ − nᵣ′)A′. Example: nᵣ = 1.51, nᵥ = 1.53: ω = 0.038.

How do I solve “Microscope and telescope: magnifying power and length” questions in JEE Main?

The focal lengths of the objective and the eyepiece and the position of the final image are given. Find the magnifying power or the length of the tube. 1. Find the objective image from the lens formula. 2. Microscope M = m₀(1 + D/fₑ) or m₀ D/fₑ. 3. Telescope M = −f₀/fₑ, L = f₀ + fₑ. Example: f₀ = 1 cm, fₑ = 5 cm, u₀ = 1.1 cm: M = 50 at infinity.

How do I solve “Defects of vision and corrective lenses” questions in JEE Main?

A person cannot see far or near objects clearly. Find the power and kind of lens needed. 1. Myopia: f = −(far point). 2. Hypermetropia: object at 25 cm, image at the near point. 3. P = 1/f in dioptre. Example: Far point 80 cm: −1.25 D; near point 50 cm: +2 D.

How do I solve “Huygens’ principle and wavefronts” questions in JEE Main?

A statement about wavefronts, secondary wavelets, or the change of speed and wavelength is made. Find the shape of the wavefront or the quantity that is unchanged. 1. Frequency is unchanged in refraction. 2. v and λ are divided by n. 3. Plane wavefront for a distant source; spherical for a point source. Example: 600 nm in glass n = 1.5: λ = 400 nm.

How do I solve “Interference of two waves: intensity and phase” questions in JEE Main?

Two waves of known intensities or amplitudes meet. Find the resultant intensity, the ratio of maximum to minimum, or the phase difference. 1. I = I₁ + I₂ + 2√(I₁I₂) cosφ. 2. Imax/Imin = ((a + 1)/(a − 1))². 3. φ = 2πΔ/λ. Example: Intensity ratio 9 : 1 gives Imax/Imin = 4.

How do I solve “Young’s double slit: fringe width and position” questions in JEE Main?

The slit separation, the screen distance and the wavelength are given. Find the fringe width, the position of a bright or dark fringe, or the effect of a medium. 1. β = λD/d. 2. Bright y = nλD/d, dark (n + ½)λD/d. 3. In a medium divide β by n. Example: λ = 600 nm, D = 1.5 m, d = 0.3 mm: β = 3 mm.

How do I solve “Slit covered by a thin sheet; two wavelengths” questions in JEE Main?

A transparent sheet covers one slit, or two wavelengths are used together. Find the shift of the fringes, the number of fringes shifted, or the position where bright fringes coincide. 1. Extra path (n − 1)t; N = (n − 1)t/λ. 2. Shift = Nβ towards the covered slit. 3. Coincidence: n₁λ₁ = n₂λ₂. Example: n = 1.5, t = 6 μm, λ = 600 nm: 5 fringes, 15 mm.

How do I solve “Single slit diffraction” questions in JEE Main?

Light passes through a narrow slit. Find the position of the first minimum, the width of the central maximum or the slit width. 1. a sinθ = nλ for minima. 2. y = Dθ for small angles. 3. Central width 2λD/a. Example: a = 0.2 mm, λ = 500 nm, D = 1 m: width 5 mm.

How do I solve “Resolving power of a telescope, microscope or eye” questions in JEE Main?

The diameter of a lens or the numerical aperture is given. Find the smallest angle or distance that can be resolved. 1. Telescope θ = 1.22λ/D. 2. Microscope d = 1.22λ/(2n sinβ). 3. Larger aperture and smaller λ resolve better. Example: 5 m mirror, 550 nm: θ = 1.34 × 10⁻⁷ rad.

How do I solve “Malus’ law and polarisers in a row” questions in JEE Main?

Unpolarised or plane polarised light falls on polarisers with axes at angles. Find the transmitted intensity or the angle needed. 1. Unpolarised through the first polariser: I₀/2. 2. Each next polariser multiplies by cos² of the angle between the axes. 3. Crossed polarisers give zero. Example: 0°, 30°, 90°: I = 3I₀/32.

How do I solve “Brewster’s angle and polarisation by reflection” questions in JEE Main?

Light falls on glass or water and the reflected ray is fully polarised. Find the angle of incidence, the index or the angle of refraction. 1. tan iʙ = n. 2. Reflected and refracted rays are perpendicular. 3. iʙ + r = 90°. Example: Glass n = 1.5: iʙ = 56.3°, r = 33.7°.

Common myths and the truth

Mistakes to avoid

Simulations for Optics

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