Reflection by spherical mirrors and the mirror equation
A shaving mirror shows a magnified face (concave) while a rear-view mirror of a car shows a wide but smaller view (convex). Link to earlier classes : reflection and image formation by mirrors from Class 10.
A spherical mirror is part of a sphere. A concave mirror converges parallel rays to its focus; a convex mirror makes them diverge as if from a focus behind it. We use the new Cartesian sign convention: all distances are measured from the pole; distances in the direction of the incident light are positive and those against it are negative. So u is negative for a real object, and f is negative for a concave mirror and positive for a convex one.
Rule 1: Mirror equation — 1 ÷ v + 1 ÷ u = 1 ÷ f and f = R ÷ 2. The magnification is m = − v ÷ u = h′ ÷ h. A negative m means an inverted image, a positive m means an erect image.
- Easy : Concave mirror, f = −15 cm, object at u = −45 cm ⇒ 1/v = −1/15 + 1/45 = −2/45, so v = −22.5 cm and m = −(−22.5) ÷ (−45) = −0.5 (real, inverted, half the size).
- Medium : The object at u = −10 cm gives 1/v = −1/15 + 1/10 = 1/30, so v = +30 cm and m = +3 (virtual, erect, three times larger, behind the mirror).
- Exam : A convex mirror with f = +20 cm and u = −30 cm gives v = +12 cm and m = +0.4: always virtual, erect and diminished.
Rule 2: Special cases — An object at the focus of a concave mirror gives an image at infinity. A convex mirror gives a virtual image for every position of the object.
- Easy : Object at u = −20 cm for a concave mirror with R = 40 cm (f = −20 cm): the object is at the focus, so v is infinite.
- Medium : Convex mirror f = +10 cm and u = −10 cm ⇒ v = +5 cm and m = +0.5.
- Exam : Mirror formula holds for both concave and convex mirrors if signs are used correctly.
In the figure : Image formed by a concave mirror when the object is beyond the centre of curvature: real, inverted and diminished.
1/v + 1/u = 1/f; f = R/2; m = −v/u; concave f < 0, convex f > 0.
Forgetting the sign of u. A real object always has a negative u in this convention.
Memory clue : "Light travels left to right: against it is negative."
Spot the difference
| Mirror | f | Image |
|---|---|---|
| Concave | negative | real or virtual, any size |
| Convex | positive | always virtual, erect, smaller |
Practice questions
- 1Write the relation between the focal length and the radius of curvature.
- 2Which mirror always forms a virtual image?
- 3What does a negative magnification show?
- 4A concave mirror has f = −15 cm and the object is at −45 cm. Find v and m.
- 5The object is moved to −10 cm. Find v and m.
- 6A convex mirror has f = +20 cm and the object is at −30 cm. Find v and m.
- 7Where is the image when the object is at the focus of a concave mirror?
- 8A convex mirror has f = +10 cm and the object is at −10 cm. Find v and m.
Show answers (try them yourself first)
- 1f = R ÷ 2.
- 2The convex mirror.
- 3The image is inverted.
- 4v = −22.5 cm, m = −0.5.
- 5v = +30 cm, m = +3 (virtual and erect).
- 6v = +12 cm, m = +0.4.
- 7At infinity.
- 8v = +5 cm, m = +0.5.
Refraction of light and Snell's law
A straw in a glass of water looks bent, and a swimming pool looks shallower than it is. Link to earlier classes : refraction and refractive index from Class 10.
Refraction is the change in the direction of light as it passes from one medium to another, caused by the change of its speed. The refractive index of a medium is n = c ÷ v. Light bends towards the normal when it enters a denser medium and away from it when it enters a rarer one.
Rule 1: Snell's law — n₁ sin i = n₂ sin r (i = angle of incidence, r = angle of refraction). The relative index is n₂₁ = n₂ ÷ n₁ = v₁ ÷ v₂.
- Easy : Air to glass (n = 1.5) at i = 30° ⇒ sin r = 0.5 ÷ 1.5 = 0.333, so r = 19.5°.
- Medium : The speed of light in water (n = 1.33) is 3 × 10⁸ ÷ 1.33 = 2.26 × 10⁸ m s⁻¹.
- Exam : Relative index of glass with respect to water = 1.5 ÷ 1.33 = 1.13.
Rule 2: Apparent depth — Seen from above, an object at real depth d in a medium of index n appears at apparent depth = d ÷ n.
- Easy : A coin at 12 cm depth in water appears at 12 ÷ 1.33 = 9.0 cm.
- Medium : A pool of 3.0 m real depth appears to be 3.0 ÷ 1.33 = 2.26 m deep.
- Exam : The frequency of light does not change on refraction; speed and wavelength change by the same factor n.
n₁ sin i = n₂ sin r; apparent depth = real depth ÷ n; n = c/v.
Measuring the angles from the surface instead of from the normal.
Memory clue : "Into dense: toward the normal. Into rare: away from it."
Spot the difference
| Medium change | Speed | Bending |
|---|---|---|
| rare to dense | decreases | towards the normal |
| dense to rare | increases | away from the normal |
Practice questions
- 1State Snell's law.
- 2Does light bend towards or away from the normal on entering glass from air?
- 3Write the formula for the refractive index in terms of speeds.
- 4Light enters glass (n = 1.5) from air at 30°. Find the angle of refraction.
- 5A coin is 12 cm deep in water (n = 1.33). Find its apparent depth.
- 6Find the speed of light in water.
- 7Find the refractive index of glass with respect to water.
- 8A pool is 3.0 m deep. How deep does it appear?
Show answers (try them yourself first)
- 1n₁ sin i = n₂ sin r.
- 2Towards the normal.
- 3n = c ÷ v.
- 4r ≈ 19.5°.
- 512 ÷ 1.33 ≈ 9.0 cm.
- 6v ≈ 2.26 × 10⁸ m s⁻¹.
- 71.5 ÷ 1.33 ≈ 1.13.
- 83.0 ÷ 1.33 ≈ 2.26 m.
Total internal reflection
A diamond sparkles because light is trapped and reflected inside it many times; optical fibres carry internet signals the same way. Link to earlier classes : refraction from the previous section.
When light goes from a denser to a rarer medium and the angle of incidence exceeds a certain limit, no light is refracted and all of it is reflected back. This is total internal reflection (TIR). The limit is the critical angle θ_c, at which the refracted ray just grazes the surface (r = 90°).
Rule 1: Critical angle — sin θ_c = 1 ÷ n (for light going from the medium to air). More generally sin θ_c = n₂ ÷ n₁ for n₁ > n₂. TIR needs (i) light travelling from the denser to the rarer medium and (ii) i greater than θ_c.
- Easy : Glass (n = 1.5) : θ_c = sin⁻¹(1 ÷ 1.5) = 41.8°; water (n = 1.33) : 48.8°.
- Medium : Diamond (n = 2.42) : θ_c = 24.4°, a small angle, so most rays entering are totally reflected and the stone sparkles.
- Exam : An optical-fibre core of n = 1.5 and cladding of n = 1.4 has θ_c = sin⁻¹(1.4 ÷ 1.5) = 69°.
Rule 2: Applications — Optical fibres (communication, endoscopes), prisms in binoculars and periscopes, mirages, and the sparkle of diamonds.
- Easy : Light stays inside a fibre by repeated TIR, however the fibre bends.
- Medium : A prism that turns light by 90° or 180° by TIR loses less light than a silvered mirror.
- Exam : TIR cannot occur when light goes from a rarer to a denser medium.
In the figure : Rays in glass meeting the surface at angles below, at and above the critical angle.
sin θ_c = 1/n; TIR only from dense to rare and for i > θ_c.
Applying TIR when light enters glass from air.
Memory clue : "Bigger n, smaller critical angle."
Spot the difference
| Angle of incidence | What happens |
|---|---|
| i < θ_c | refraction and weak reflection |
| i = θ_c | refracted ray grazes the surface |
| i > θ_c | total internal reflection |
Practice questions
- 1State the two conditions for total internal reflection.
- 2Write the formula for the critical angle.
- 3Name one application of TIR.
- 4Find the critical angle for glass of n = 1.5.
- 5Find it for water of n = 1.33.
- 6Find it for diamond of n = 2.42.
- 7A fibre has a core of n = 1.5 and a cladding of n = 1.4. Find the critical angle at the core–cladding boundary.
- 8Why does a diamond sparkle more than glass?
Show answers (try them yourself first)
- 1Light goes from a denser to a rarer medium, and the angle of incidence is greater than the critical angle.
- 2sin θ_c = 1 ÷ n.
- 3Optical fibre (or prism binoculars, or an endoscope).
- 4θ_c ≈ 41.8°.
- 5θ_c ≈ 48.8°.
- 6θ_c ≈ 24.4°.
- 7θ_c = sin⁻¹(1.4 ÷ 1.5) ≈ 69°.
- 8Its refractive index is high, so its critical angle is small and much light is totally reflected inside.
Refraction at a spherical surface and by a thin lens
A magnifying glass, a camera lens and the lens of the eye all form images by refraction at curved surfaces. Link to earlier classes : lenses and lens formula from Class 10.
A thin lens is bounded by two spherical surfaces. A convex lens converges light (f positive); a concave lens diverges it (f negative). We use the same Cartesian sign convention, with distances measured from the optical centre.
Rule 1: Spherical surface and lens maker's formula — For one surface, n₂ ÷ v − n₁ ÷ u = (n₂ − n₁) ÷ R. For a thin lens of index n in air, 1 ÷ f = (n − 1)(1 ÷ R₁ − 1 ÷ R₂).
- Easy : Air to glass (n = 1.5), R = +20 cm, u = −60 cm ⇒ 1.5/v = 0.5/20 − 1/60 = 1/120, so v = 180 cm.
- Medium : A double-convex lens (R₁ = +20, R₂ = −20 cm, n = 1.5) has 1/f = 0.5 × (1/20 + 1/20) = 1/20, so f = 20 cm.
- Exam : A plano-convex lens with n = 1.5 and R = 15 cm has f = 30 cm.
Rule 2: Thin lens formula — 1 ÷ v − 1 ÷ u = 1 ÷ f and m = v ÷ u.
- Easy : f = +20 cm and u = −30 cm ⇒ 1/v = 1/20 − 1/30 = 1/60, so v = +60 cm and m = −2 (real, inverted, twice as large).
- Medium : A concave lens with f = −15 cm and u = −30 cm gives 1/v = −1/15 − 1/30 = −1/10, so v = −10 cm and m = +1/3 (virtual, erect, smaller).
- Exam : A concave lens always gives a virtual, erect, smaller image.
In the figure : A convex lens forming a real, inverted image of an object placed beyond its focus.
1/v − 1/u = 1/f; m = v/u; 1/f = (n−1)(1/R₁ − 1/R₂).
Using the mirror form 1/v + 1/u for a lens. A lens has a minus sign.
Memory clue : "Lens: v minus u. Mirror: v plus u."
Spot the difference
| Lens | f | Image |
|---|---|---|
| Convex | positive | real or virtual |
| Concave | negative | always virtual, erect, smaller |
Practice questions
- 1Write the lens formula.
- 2What is the sign of f for a concave lens?
- 3Write the lens maker's formula.
- 4Find v for refraction at a convex surface (R = +20 cm, air to glass n = 1.5) of an object at −60 cm.
- 5Find f of a double-convex lens with R₁ = +20 cm, R₂ = −20 cm and n = 1.5.
- 6A convex lens with f = 20 cm has an object at −30 cm. Find v and m.
- 7A concave lens has f = −15 cm and the object is at −30 cm. Find v and m.
- 8Find f of a plano-convex lens with R = 15 cm and n = 1.5.
Show answers (try them yourself first)
- 11/v − 1/u = 1/f.
- 2Negative.
- 31/f = (n − 1)(1/R₁ − 1/R₂).
- 4v = 180 cm.
- 5f = 20 cm.
- 6v = +60 cm, m = −2.
- 7v = −10 cm, m = +1/3.
- 81/f = 0.5 × (1/15), so f = 30 cm.
Power of a lens and lenses in contact
An optician writes the power of a spectacle lens in dioptre, such as +2.0 D or −1.5 D. Link to earlier classes : power of a lens from Class 10.
The power of a lens is the reciprocal of its focal length in metres. A convex lens has positive power, a concave lens negative. The unit is the dioptre (D), 1 D = 1 m⁻¹.
Rule 1: Power — P = 1 ÷ f (f in metres). A short focal length means a strong lens.
- Easy : f = 25 cm = 0.25 m ⇒ P = +4 D.
- Medium : A lens of −2 D has f = −0.5 m = −50 cm.
- Exam : Doctors prescribe spectacles by power, not by focal length.
Rule 2: Lenses in contact — 1 ÷ f = 1 ÷ f₁ + 1 ÷ f₂, so P = P₁ + P₂. The magnification of the combination is the product of the individual magnifications.
- Easy : +5 D and −2 D in contact give +3 D, that is f = 33 cm.
- Medium : Convex lenses of 40 cm and 60 cm in contact have 1/f = 1/40 + 1/60, so f = 24 cm and P = 4.2 D.
- Exam : A convex lens of 20 cm and a concave lens of 30 cm in contact have 1/f = 1/20 − 1/30, so f = +60 cm.
P = 1/f (metres); P = P₁ + P₂ in contact; unit dioptre.
Using f in centimetres in the formula for power in dioptre.
Memory clue : "Convert cm to m first: P is one over f in metres."
Spot the difference
| Quantity | Sign for convex | Sign for concave |
|---|---|---|
| f | + | − |
| P | + | − |
Practice questions
- 1Write the unit of the power of a lens.
- 2What is the sign of the power of a concave lens?
- 3How do powers combine for lenses in contact?
- 4Find the power of a lens of f = 25 cm.
- 5Find the focal length of a −2 D lens.
- 6Find the power and focal length of +5 D and −2 D lenses in contact.
- 7A convex lens of 20 cm and a concave lens of 30 cm are in contact. Find the focal length of the combination.
- 8Two convex lenses of 40 cm and 60 cm are in contact. Find the combined focal length.
Show answers (try them yourself first)
- 1Dioptre (D = m⁻¹).
- 2Negative.
- 3They add: P = P₁ + P₂.
- 4P = 1 ÷ 0.25 = +4 D.
- 5f = −0.5 m = −50 cm.
- 6P = +3 D, f ≈ 33 cm.
- 71/f = 1/20 − 1/30 = 1/60, so f = +60 cm.
- 81/f = 1/40 + 1/60 = 1/24, so f = 24 cm.
Refraction through a prism
A glass prism splits sunlight into the colours of a rainbow. Link to earlier classes : dispersion of light from Class 10 and Snell's law from this chapter.
A prism bends light towards its base. The angle between the incident ray and the emergent ray is the angle of deviation δ. As the angle of incidence is increased, δ first decreases to a least value, the angle of minimum deviation D, and then increases. Because the refractive index depends on colour, a prism also disperses white light.
Rule 1: Deviation — δ = i + e − A (i = incidence angle, e = emergent angle, A = angle of the prism). At minimum deviation i = e and n = sin[(A + D) ÷ 2] ÷ sin(A ÷ 2).
- Easy : i = 40°, e = 50° and A = 60° ⇒ δ = 90° − 60° = 30°.
- Medium : For A = 60° and n = 1.5, sin[(60° + D)/2] = 0.75, so (60° + D)/2 = 48.6° and D = 37.2°.
- Exam : From D = 37.2° and A = 60°, n = sin 48.6° ÷ sin 30° = 0.75 ÷ 0.5 = 1.5.
Rule 2: Thin prism and dispersion — For a thin prism, δ = (n − 1) A. Violet has a higher n than red, so violet is deviated more.
- Easy : A = 4° and n = 1.5 ⇒ δ = 0.5 × 4° = 2°.
- Medium : In a rainbow the colour at the top is red and the one at the bottom is violet.
- Exam : Red light is deviated least and violet most, which makes the dispersion of white light into a spectrum.
δ = i + e − A; at minimum deviation i = e; thin prism δ = (n−1)A.
Thinking that deviation increases steadily with the angle of incidence.
Memory clue : "Violet bends the most, red the least: VIBGYOR from the base."
Spot the difference
| Colour | n | Deviation |
|---|---|---|
| Red | smallest | least |
| Violet | largest | most |
Practice questions
- 1Write the formula for the angle of deviation in a prism.
- 2What is the relation between i and e at minimum deviation?
- 3Which colour is deviated most by a prism?
- 4A prism with A = 60° has n = 1.5. Find the angle of minimum deviation.
- 5i = 40°, e = 50° and A = 60°. Find δ.
- 6A thin prism has A = 4° and n = 1.5. Find δ.
- 7Find n from D = 37.2° and A = 60°.
- 8Why does a prism disperse white light?
Show answers (try them yourself first)
- 1δ = i + e − A.
- 2i = e.
- 3Violet.
- 4D ≈ 37.2°.
- 5δ = 30°.
- 6δ = 2°.
- 7n = sin 48.6° ÷ sin 30° ≈ 1.5.
- 8The refractive index of glass is different for different colours, so each colour is deviated by a different amount.
Optical instruments: microscope and telescope
A microscope shows tiny cells and a telescope shows distant craters of the Moon. Link to earlier classes : lenses from the earlier sections and the human eye from Class 10.
The magnifying power of an instrument is the ratio of the angle seen with the instrument to the angle seen without it. A compound microscope has two convex lenses, an objective of short focal length near the object and an eyepiece near the eye. An astronomical telescope has an objective of long focal length and an eyepiece of short focal length.
Rule 1: Simple and compound microscope — A simple microscope (magnifier) has m = 1 + D ÷ f with the image at the near point (D = 25 cm), and m = D ÷ f with the image at infinity. For a compound microscope, m ≈ (L ÷ f_o) × (D ÷ f_e) (L = distance between the lenses).
- Easy : A magnifier with f = 5 cm gives m = 1 + 25/5 = 6 (image at the near point) or 5 (image at infinity).
- Medium : With L = 15 cm, f_o = 1 cm, f_e = 5 cm and D = 25 cm, m = (15 ÷ 1) × (25 ÷ 5) = 75.
- Exam : A short f_o and f_e give a large magnification.
Rule 2: Astronomical telescope — In normal adjustment (final image at infinity) m = f_o ÷ f_e and the tube length is f_o + f_e. A large objective collects more light and gives better resolution.
- Easy : f_o = 100 cm and f_e = 2.5 cm ⇒ m = 40.
- Medium : The length of the tube is 100 + 2.5 = 102.5 cm.
- Exam : Large telescopes use mirrors as objectives (reflecting telescopes) because large mirrors are easier to make and support, and they avoid colour defects.
Microscope: m = (L/f_o)(D/f_e); telescope: m = f_o/f_e.
Swapping the roles: the telescope's objective has the longer focal length, the microscope's the shorter.
Memory clue : "Telescope: Objective long. Microscope: Objective tiny."
Spot the difference
| Instrument | Objective | Eyepiece |
|---|---|---|
| Microscope | short f | longer f |
| Telescope | long f | short f |
Practice questions
- 1Define magnifying power of an optical instrument.
- 2In a telescope, which lens has the larger focal length?
- 3What is the position of the final image in the normal adjustment of a telescope?
- 4Find the magnifying power of a compound microscope with L = 15 cm, f_o = 1 cm, f_e = 5 cm and D = 25 cm.
- 5Find the magnifying power of a telescope with f_o = 100 cm and f_e = 2.5 cm.
- 6Find the length of that telescope.
- 7Find the magnifying power of a magnifier of f = 5 cm with the image at the near point and at infinity.
- 8Why is a large aperture of the objective useful in a telescope?
Show answers (try them yourself first)
- 1The ratio of the angle subtended by the image to that of the object seen directly.
- 2The objective.
- 3At infinity.
- 4m = 75.
- 5m = 40.
- 6102.5 cm.
- 76 at the near point and 5 at infinity.
- 8It collects more light (brighter image) and gives better resolution.