Reflection at Mirrors
Laws of Reflection
The angle of incidence equals the angle of reflection, and the incident ray, reflected ray, and normal all lie in the same plane — these laws hold for both plane and curved mirrors.
Concave & Convex Mirrors
A concave mirror (reflecting surface curves inward) can form real, inverted images (for objects beyond the focus) or virtual, magnified images (for objects within the focus) — used in torches, headlights, and shaving mirrors. A convex mirror always forms a virtual, diminished, upright image — used in vehicle side mirrors for a wider field of view.
Solved Example
Explain why convex mirrors, rather than concave or plane mirrors, are used as vehicle side mirrors.
A convex mirror always produces a virtual, upright, diminished image regardless of the object's distance — this diminished image effectively represents a much wider field of view compared to a plane mirror of the same size, allowing the driver to see a larger area behind and to the side of the vehicle, which is crucial for safety despite the slight distortion in perceived distance ("objects in mirror are closer than they appear").
Q. Why can a concave mirror produce either a real or a virtual image, depending on the object's position, while a convex mirror always produces only a virtual image?
Ans: A concave mirror converges reflected rays — when the object is beyond the focus, these converging rays actually meet at a point in front of the mirror, forming a real image; when the object is within the focus, the reflected rays diverge after reflection and only appear to meet behind the mirror, forming a virtual image. A convex mirror always diverges reflected rays regardless of object position, so the rays never actually converge in front of the mirror, meaning only a virtual image (where rays appear to meet behind the mirror) can ever be formed.
Q. Why do dentists commonly use concave mirrors to examine teeth?
Ans: When an object (a tooth) is placed within the focal length of a concave mirror, the mirror produces a virtual, upright, and magnified image — this magnification allows the dentist to see fine details of the tooth's surface much more clearly than would be possible with the naked eye or a plane mirror.
Mirror Formula & Magnification
Magnification
$$m = \dfrac{h_i}{h_o} = -\dfrac{v}{u}$$
Solved Example
An object is placed 30 cm from a concave mirror of focal length 10 cm. Calculate the image distance and magnification.
Using sign convention, $u = -30\ \text{cm}$, $f = -10\ \text{cm}$: $\dfrac{1}{v} = \dfrac{1}{f}-\dfrac{1}{u} = \dfrac{1}{-10}-\dfrac{1}{-30} = -\dfrac{1}{10}+\dfrac{1}{30} = \dfrac{-3+1}{30} = \dfrac{-2}{30}$
$v = -15\ \text{cm}$ (real image, in front of the mirror)
$m = -\dfrac{v}{u} = -\dfrac{-15}{-30} = -0.5$ (inverted, diminished image)
Q. Why does a negative value of magnification indicate an inverted image, while a positive value indicates an upright image?
Ans: Magnification is defined as the ratio of image height to object height (with sign), m = h_i/h_o — by convention, heights measured upward (above the principal axis) are taken as positive and downward as negative. A negative magnification means the image height has the opposite sign to the object height, meaning the image is inverted relative to the object; a positive magnification means both share the same sign, meaning the image is upright, in the same orientation as the object.
Q. An object placed 20 cm from a convex mirror of focal length 15 cm forms an image. Calculate the image distance.
Ans: For a convex mirror, f = +15 cm, u = −20 cm. 1/v = 1/f − 1/u = 1/15 − 1/(−20) = 1/15 + 1/20 = (4+3)/60 = 7/60. v = 60/7 ≈ 8.57 cm (positive, confirming a virtual image behind the mirror).
Refraction & Snell's Law
Refraction
Refraction is the bending of light as it passes from one medium into another with a different optical density, due to a change in the speed of light in the new medium.
Solved Example
Light travels from air (n=1) into glass (n=1.5) at an angle of incidence of 30°. Calculate the angle of refraction.
$n_1\sin\theta_1 = n_2\sin\theta_2 \Rightarrow (1)\sin30° = (1.5)\sin\theta_2 \Rightarrow \sin\theta_2 = \dfrac{0.5}{1.5} = 0.333$
$\theta_2 = \sin^{-1}(0.333) \approx 19.5°$
Q. Why does light bend towards the normal when entering a denser medium (like glass) from a rarer medium (like air)?
Ans: Light travels more slowly in a denser (higher refractive index) medium. As a wavefront enters the denser medium at an angle, the portion of the wavefront that enters first begins slowing down while the rest of the wavefront (still in the rarer medium) continues at its original, faster speed — this differential change in speed across the wavefront causes it to bend, refracting the ray's direction towards the normal in the denser medium.
Q. Light travels from water (n=1.33) into air at an angle of incidence of 20°. Calculate the angle of refraction.
Ans: (1.33)sin20° = (1)sinθ₂ → sinθ₂ = 1.33 × 0.342 = 0.455 → θ₂ = sin⁻¹(0.455) ≈ 27.1°.
Apparent Depth
Due to refraction, an object submerged in a denser medium (like water) appears shallower (closer to the surface) than its actual depth, when viewed from a rarer medium (like air) directly above: $\dfrac{\text{Real depth}}{\text{Apparent depth}} = n$ (refractive index of the denser medium).
Solved Example
A coin lies at the bottom of a pool of water (n=1.33) with an actual depth of 2 m. Calculate its apparent depth when viewed from directly above.
$\text{Apparent depth} = \dfrac{\text{Real depth}}{n} = \dfrac{2}{1.33} \approx 1.5\ \text{m}$
Q. Why does a swimming pool appear shallower than its actual depth when viewed from above the water's surface?
Ans: Light rays from the pool's bottom refract (bend away from the normal) as they exit the denser water into the rarer air above — to an observer looking down, these refracted rays appear to originate from a point higher up (shallower) than the object's actual position, since our brain assumes light travels in straight lines back to its apparent source, causing the pool to look shallower than it truly is.
Total Internal Reflection
Definition
Total internal reflection (TIR) occurs when light travelling from a denser to a rarer medium strikes the boundary at an angle greater than the critical angle — beyond this angle, no refraction occurs at all, and all the light is completely reflected back into the denser medium.
Critical Angle
$$\sin\theta_c = \dfrac{n_2}{n_1} \ \ (\text{for } n_1 > n_2)$$
Solved Example
Calculate the critical angle for light travelling from glass (n=1.5) into air (n=1).
$\sin\theta_c = \dfrac{n_2}{n_1} = \dfrac{1}{1.5} = 0.667 \Rightarrow \theta_c = \sin^{-1}(0.667) \approx 41.8°$
Q. Why can total internal reflection only occur when light travels from a denser medium to a rarer medium, and not the other way around?
Ans: Total internal reflection relies on the refracted ray bending away from the normal so much (as the angle of incidence increases) that it eventually would need to bend to 90° or beyond — this only happens when going from a denser to a rarer medium (where the refracted ray bends away from the normal). Going from a rarer to a denser medium, the refracted ray always bends towards the normal, and can never reach or exceed 90°, so there's no critical angle at which total reflection would occur in that direction.
Q. How does an optical fibre use total internal reflection to transmit light signals over long distances with minimal loss?
Ans: An optical fibre consists of a core with a higher refractive index surrounded by a cladding of lower refractive index. Light entering the fibre at a suitable angle repeatedly strikes the core-cladding boundary at an angle greater than the critical angle, undergoing total internal reflection each time — this confines the light entirely within the core, allowing it to travel long distances through repeated total internal reflections with very little loss of intensity, since virtually none of the light escapes through the boundary.
Solved Example
Explain how the phenomenon of a mirage (appearing as a shimmering pool of water on a hot road) is caused by total internal reflection.
On a hot day, air just above the road surface becomes much hotter (and hence less dense, lower refractive index) than the cooler air higher up. Light from the sky travelling towards the road passes through layers of progressively hotter, less dense air, continuously refracting and bending — eventually, the light can bend so much that it undergoes something similar to total internal reflection at a shallow angle near the hot road surface, curving back upward towards an observer's eyes. This creates the illusion of a reflective, water-like surface on the road, even though no actual water or reflecting surface is present — purely an effect of refraction through non-uniform air layers.
Lens Formula & Lens Maker's Formula
Lens Formula
$$\dfrac{1}{v}-\dfrac{1}{u} = \dfrac{1}{f}$$
Solved Example
An object is placed 20 cm from a convex lens of focal length 10 cm. Calculate the image distance.
Using sign convention, $u = -20\ \text{cm}$, $f = +10\ \text{cm}$: $\dfrac{1}{v} = \dfrac{1}{f}+\dfrac{1}{u} = \dfrac{1}{10}+\dfrac{1}{-20} = \dfrac{2-1}{20} = \dfrac{1}{20}$
$v = 20\ \text{cm}$ (real, inverted image on the opposite side)
Q. Why does a convex lens act as a converging lens, while a concave lens acts as a diverging lens?
Ans: A convex lens is thicker at the centre than at the edges — light rays passing through the thicker central region are refracted (bent) more strongly towards the axis than rays passing through the thinner edge regions, causing parallel rays to converge towards a common focal point. A concave lens is thinner at the centre than at the edges, causing the opposite effect — rays are bent away from the axis, making parallel rays diverge as if originating from a common focal point on the same side as the incoming light.
Q. A convex lens has radii of curvature 20 cm and 30 cm (with the standard sign convention giving R₁ = +20 cm, R₂ = −30 cm), and refractive index 1.5. Calculate its focal length.
Ans: 1/f = (n−1)(1/R₁ − 1/R₂) = (0.5)(1/20 − 1/(−30)) = (0.5)(1/20+1/30) = (0.5)(3/60+2/60) = (0.5)(5/60) = 1/24. So f = 24 cm.
Power of a Lens & Combination of Lenses
Power of a Lens
The power of a lens is a measure of its ability to converge or diverge light, defined as the reciprocal of its focal length (in metres): $P = 1/f$, measured in dioptres (D). A converging (convex) lens has positive power; a diverging (concave) lens has negative power.
Combination of Lenses in Contact
For lenses placed in close contact, the powers (and hence reciprocals of focal length) simply add: $P = P_1+P_2+\cdots$, equivalently $\dfrac{1}{f} = \dfrac{1}{f_1}+\dfrac{1}{f_2}+\cdots$
Solved Example
Two lenses of power +5 D and −2 D are placed in contact. Calculate the power and focal length of the combination.
$P = P_1+P_2 = 5+(-2) = 3\ \text{D}$
$f = \dfrac{1}{P} = \dfrac{1}{3}\ \text{m} \approx 33.3\ \text{cm}$
Q. Why is power, rather than focal length itself, used when combining multiple lenses in contact?
Ans: Power is directly proportional to a lens's converging/diverging ability, and this ability adds in a simple, linear fashion when lenses are placed close together (each lens acts on the light after it has already been refracted by the previous lens) — this direct additivity makes power (1/f) mathematically much simpler to combine than focal length values themselves, which don't add in an equally straightforward way.
Q. A concave lens has a focal length of −25 cm. Calculate its power in dioptres.
Ans: P = 1/f = 1/(−0.25 m) = −4 D.
Prism & Dispersion
Refraction Through a Prism
Light passing through a prism is refracted twice (entering and exiting), causing an overall deviation from its original path — the angle of deviation depends on the prism's refracting angle, refractive index, and the angle of incidence.
Dispersion
Dispersion is the splitting of white light into its constituent colours (a spectrum) upon passing through a prism, because different wavelengths (colours) of light have slightly different refractive indices in the same material — violet light bends the most (highest refractive index), and red light bends the least.
Solved Example
Explain why violet light is deviated more than red light when white light passes through a glass prism.
The refractive index of glass (and most transparent materials) varies slightly with wavelength — shorter wavelengths (like violet) generally experience a higher refractive index than longer wavelengths (like red), a phenomenon called dispersion. Since a larger refractive index causes greater bending (deviation) of light passing through the prism, violet light — having the higher refractive index — is deviated more from its original path than red light, causing the spreading out of white light into its familiar rainbow of colours upon exiting the prism.
Q. Why does a rainbow form when sunlight passes through raindrops, in terms of the same dispersion principle seen in a prism?
Ans: Each raindrop acts like a tiny natural prism combined with an internal reflecting surface — sunlight entering a raindrop is refracted (with different colours bending by slightly different amounts, exactly as in a prism), undergoes total internal reflection off the back of the droplet, and is refracted again upon exiting, further separating the different colours — the combined refraction-reflection-refraction process across countless raindrops, viewed from the correct angle, produces the distinct, separated bands of colour that make up the visible rainbow.
Q. Why does a prism not simply reflect all colours of light equally, but instead separate them into a spectrum?
Ans: A prism relies on refraction, not simple reflection, to bend light — and since the refractive index of the prism's material genuinely differs (even if only slightly) for different wavelengths of light, each colour bends by a correspondingly different amount as it passes through the prism, causing them to separate into distinct paths (a spectrum) rather than continuing together as combined white light.
Microscopes
Simple Microscope (Magnifying Glass)
A single convex lens used to view a small object placed within its focal length, producing a virtual, magnified image. Angular magnification (viewing at the near point, $D=25\ \text{cm}$): $m = 1+\dfrac{D}{f}$.
Compound Microscope
Uses two converging lenses — an objective lens (short focal length, close to the object, forms a real, magnified image) and an eyepiece (acts as a simple magnifier on this intermediate image, producing a final large virtual image) — achieving much greater total magnification than a single lens alone.
Solved Example
Calculate the angular magnification of a simple microscope (magnifying glass) with focal length 5 cm, for a person with a near point of 25 cm.
$m = 1+\dfrac{D}{f} = 1+\dfrac{25}{5} = 1+5 = 6$
Q. Why does a compound microscope achieve significantly greater total magnification than either of its two individual lenses could achieve alone?
Ans: The overall magnification of a compound microscope is the product of the magnifications produced by the objective lens and the eyepiece individually (total magnification = objective magnification × eyepiece magnification) — since each lens contributes its own multiplicative magnifying effect on the image already magnified by the previous lens, the combined, sequential effect is considerably greater than what either single lens could produce working alone.
Q. Why is the objective lens of a compound microscope designed with a very short focal length?
Ans: A shorter focal length for the objective lens allows the object to be placed very close to the lens (just beyond its focal point) while still producing a highly magnified, real intermediate image — this significant initial magnification from the objective is essential for achieving the very high overall magnification a compound microscope is designed to provide, before the eyepiece further magnifies this already-enlarged intermediate image.
Astronomical Telescope
Principle
An astronomical (refracting) telescope uses two converging lenses — an objective lens with a large focal length to collect light from distant objects, and an eyepiece with a short focal length to further magnify the resulting image — designed to increase the apparent angular size of distant objects, not to magnify their linear size (unlike a microscope).
Solved Example
An astronomical telescope has an objective lens of focal length 100 cm and an eyepiece of focal length 5 cm. Calculate its magnifying power.
$m = \dfrac{f_o}{f_e} = \dfrac{100}{5} = 20$
Q. Why does a telescope's objective lens have a large focal length, while its eyepiece has a small focal length — the opposite arrangement to a compound microscope?
Ans: A telescope needs to gather as much light as possible from very distant (and often faint) objects and form a reasonably sized initial image — a large focal length objective achieves this while keeping aberrations manageable for a wide aperture. The eyepiece then needs to provide strong additional angular magnification of this already-formed image, which requires a short focal length. A microscope, by contrast, deals with nearby, small objects, requiring the objective itself to have a short focal length to achieve strong initial magnification of a tiny, close object.
Q. If a telescope's eyepiece is replaced with one of a much shorter focal length (objective unchanged), how does its magnifying power change?
Ans: Since m = f_o/f_e, decreasing f_e (with f_o fixed) increases the magnifying power m.
Young's Double Slit Experiment & Interference
Young's Double Slit Experiment
Light passing through two closely spaced, narrow slits produces an interference pattern of alternating bright and dark fringes on a screen — direct experimental evidence for the wave nature of light, since particle-based models of light cannot explain this pattern of constructive and destructive interference.
Solved Example
In a Young's double slit experiment, the slit separation is 0.5 mm, the screen distance is 1 m, and the wavelength of light used is 500 nm. Calculate the fringe width.
$\beta = \dfrac{\lambda D}{d} = \dfrac{(500\times10^{-9})(1)}{0.5\times10^{-3}} = \dfrac{5\times10^{-7}}{5\times10^{-4}} = 1\times10^{-3}\ \text{m} = 1\ \text{mm}$
Q. Why does Young's double slit experiment provide such compelling evidence for the wave nature of light?
Ans: The alternating pattern of bright and dark fringes observed can only be explained by constructive interference (where waves from the two slits arrive in phase, reinforcing each other) and destructive interference (where waves arrive out of phase, cancelling each other) — this is a distinctly wave-like phenomenon. A simple particle model of light, where light particles would just travel in straight lines and pile up directly behind each slit, cannot account for this specific pattern of alternating brightness and darkness, making the interference pattern strong, direct evidence that light behaves as a wave.
Q. If the slit separation in a Young's double slit setup is doubled (wavelength and screen distance unchanged), how does the fringe width change?
Ans: Since β ∝ 1/d, doubling the slit separation halves the fringe width.
Diffraction
Definition
Diffraction is the bending/spreading of waves as they pass around obstacles or through narrow openings — most noticeable when the size of the obstacle/opening is comparable to the wavelength of the wave involved.
Diffraction vs Interference — Key Distinction
Interference (Young's double slit) arises from the superposition of waves from two or more distinct, separate sources. Diffraction arises from the superposition of wavelets from different parts of the same single wavefront passing through an aperture or around an obstacle — a subtle but important conceptual distinction, even though both phenomena rely on the same underlying principle of superposition.
Solved Example
Explain why diffraction effects are much more noticeable for sound waves passing through an open doorway than for light waves passing through the same doorway.
Diffraction effects become significant when the size of the opening is comparable to (or smaller than) the wavelength of the wave. Sound waves have wavelengths typically on the order of centimetres to metres — comparable to or larger than a typical doorway's width — so sound diffracts (spreads out) substantially as it passes through, allowing it to be heard clearly even around corners. Light waves have an extremely tiny wavelength (a few hundred nanometres) compared to a doorway's size, so diffraction effects for light through a doorway are negligible, and light essentially appears to travel in straight lines (rays) through such a large opening.
Q. Why is diffraction generally more noticeable for waves passing through a narrow slit, compared to a very wide opening?
Ans: Diffraction effects become significant specifically when the opening's size is comparable to or smaller than the wave's wavelength — a narrow slit satisfies this condition much more readily than a wide opening does, causing pronounced spreading/bending of the wave as it passes through. A very wide opening (much larger than the wavelength) allows the wave to pass through largely undisturbed, behaving more like a simple, undiffracted beam continuing in its original direction.
Q. What is the key conceptual difference between interference and diffraction, even though both involve the superposition of waves?
Ans: Interference involves the superposition of waves originating from two or more distinct, separate sources (like the two slits in Young's experiment). Diffraction involves the superposition of secondary wavelets originating from different parts of a single wavefront as it passes through or around an obstacle/opening — the distinction lies in whether the interfering waves come from genuinely separate sources or from different portions of the same original wavefront.
Polarization
Definition
Polarization is the phenomenon of restricting the oscillation of a transverse wave (like light) to a single specific plane — unpolarised light (like ordinary sunlight) oscillates in all possible perpendicular directions; a polarising filter selectively transmits only the component oscillating along its own specific transmission axis.
Solved Example
Polarised light of intensity 100 W/m² passes through an analyser whose transmission axis is at 60° to the polarisation direction. Calculate the transmitted intensity.
$I = I_0\cos^2\theta = 100\cos^2(60°) = 100(0.5)^2 = 100(0.25) = 25\ \text{W/m}^2$
Q. Why does rotating a polarising filter placed in front of already-polarised light cause the transmitted intensity to vary, reaching zero at a specific orientation?
Ans: By Malus's law, transmitted intensity depends on cos²θ, where θ is the angle between the incoming light's polarisation direction and the filter's transmission axis. As the filter is rotated, θ changes continuously, causing the transmitted intensity to vary correspondingly — when θ reaches exactly 90° (filter's transmission axis perpendicular to the light's polarisation direction), cos²90° = 0, meaning no light at all passes through, explaining why rotating a polariser in front of polarised light can completely block it at this specific orientation.
Q. Why are polarised sunglasses particularly effective at reducing glare from reflected light off surfaces like water or wet roads?
Ans: Light reflected off a horizontal surface (like water or a wet road) tends to become significantly polarised in the horizontal direction upon reflection. Polarised sunglasses are specifically designed with their transmission axis oriented vertically, blocking most of this horizontally-polarised, glare-causing reflected light while still allowing regular, unpolarised ambient light (which has components in all directions, including vertical) to pass through relatively unaffected — significantly reducing glare without proportionally darkening the overall scene.
⚡ Formula Cheat-Sheet
Mirror formula1/v + 1/u = 1/f
Magnification (mirror)m = −v/u
Snell's lawn₁sinθ₁ = n₂sinθ₂
Critical anglesinθc = n₂/n₁
Lens formula1/v − 1/u = 1/f
Power of lensP = 1/f
Simple microscope magnificationm = 1 + D/f
Telescope magnificationm = f₀/f_e
Fringe width (YDSE)β = λD/d
Malus's lawI = I₀cos²θ