Optics

You cover the lower half of a camera lens with black tape. What happens to the photograph?

Most say the bottom half of the picture disappears.

The whole image stays, just dimmer. Every point of the lens sends light to every point of the image, so blocking part of the aperture removes brightness, never area.

The largest chapter in the syllabus, and the one most often studied as two unrelated halves. Three facts join them:

• Ray and wave optics are two limits of one theory. Obstacles far larger than the wavelength give straight lines; obstacles comparable to it give diffraction. Light's 500 nm wavelength is why everyday objects sit firmly in the ray limit.
• One sign convention turns every mirror and lens problem into the same problem. Measure from the pole or optical centre, take the incident direction as positive, and apply it without exception — even when a distance is obviously going to come out negative.
• Interference and diffraction redistribute energy, never create it. Total light on the screen is unchanged; only its arrangement differs.

Scope note. Resolving power of microscopes and telescopes was removed from JEE Main in the 2023 revision and stays out for 2026. Prisms, dispersion and everything else here is still in.

1. Reflection and Spherical Mirrors

Incident ray, reflected ray and normal lie in one plane, and $\angle i=\angle r$.

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f},f=\frac{R}{2},m=-\frac{v}{u}$

A concave mirror can form real or virtual images. A convex mirror always gives a virtual, erect, diminished image, whatever the object position.

Negative magnification means an inverted image, and from a single mirror or lens an inverted image is always real.

Image positions for a concave mirror

Only an object inside the focal length gives a virtual erect image — which is why a shaving mirror has to be held close to the face.

Object and image swap along the same paths. Put the object where the image was and the image lands where the object was: reversibility, and a fast way to check an answer.

Illustration 1

A concave mirror of focal length 20 cm forms an image twice the size of the object. Find the object distance.

There are two answers, because $|m|=2$ allows $m=-2$ (real, inverted) or $m=+2$ (virtual, erect).

Real case, $m=-v/u=-2\Rightarrow v=2u$:

$\frac{1}{2u}+\frac{1}{u}=\frac{1}{-20}\Rightarrow \frac{3}{2u}=-\frac{1}{20}\Rightarrow u=-30\text{cm}$

Virtual case, $m=+2\Rightarrow v=-2u$:

$-\frac{1}{2u}+\frac{1}{u}=\frac{1}{-20}\Rightarrow \frac{1}{2u}=-\frac{1}{20}\Rightarrow u=-10\text{cm}$

30 cm (outside $F$) or 10 cm (inside $F$). Quoting only one is the standard way to lose the mark.

2. The Sign Convention

Measure from the pole of the mirror or optical centre of the lens; take the direction of incident light as positive.

Trap. Apply the convention mechanically and let the algebra deliver the signs. Deciding in advance whether an image "should" be real is what causes errors. A positive $v$ for a mirror means in front and real; a negative $v$ means behind and virtual — the opposite of the lens case.

3. Refraction at Plane Surfaces

$n_1\sin {\theta }_1=n_2\sin {\theta }_2,n=\frac{c}{v}$

Entering a denser medium, light bends toward the normal and slows. Frequency is fixed by the source; wavelength shortens by $n$.

The slab shifts without magnifying because the emergent ray is parallel to the incident one — displaced, not deviated.

Illustration 2

A coin lies at the bottom of a beaker holding 12 cm of water ($n=4/3$). A glass slab 3 cm thick ($n=1.5$) is laid on the surface. Find the total apparent shift.

Each layer contributes independently:

$\text{water: }12\left(1-\frac{3}{4}\right)=3\text{cm},\text{glass: }3\left(1-\frac{1}{1.5}\right)=1\text{cm}$

$\text{total shift}=4\text{cm}$

Layers add. The coin appears 4 cm nearer than it is, and this stacking is how a real optical path through several media is handled.

4. Total Internal Reflection

$\sin C=\frac{1}{n}$

Beyond the critical angle, light going from denser to rarer does not emerge at all.

Total internal reflection is the only perfect reflection in physics. An ordinary mirror always absorbs a few per cent; TIR loses nothing.

Illustration 3

An optical fibre has core index 1.50 and cladding 1.48. Find the critical angle at the core-cladding boundary, and the largest angle at the flat end face that still gets guided.

$\sin C=\frac{1.48}{1.50}=0.9867\Rightarrow C=80.6^{\circ }$

A ray must strike the wall at more than $80.6^{\circ }$, so it must travel almost parallel to the axis. The acceptance cone at the entrance follows:

$\text{NA}=\sqrt{n_1^2-n_2^2}=\sqrt{2.25-2.1904}=0.244\Rightarrow {\theta }_{max}=14.1^{\circ }$

The cladding is barely different from the core, and that near-equality is deliberate: it keeps the acceptance cone narrow, so all guided rays take nearly the same path length and a pulse does not smear out over kilometres.

5. Spherical Surfaces and Lenses

$\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}$

Apply it once at each surface of a thin lens and the lens maker's formula falls out:

$\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right),\frac{1}{v}-\frac{1}{u}=\frac{1}{f},m=\frac{v}{u}$

Trap. The lens formula has a minus where the mirror formula has a plus, and the magnification loses its minus sign. Mixing them is a routine source of lost marks.

$n$ here is the index relative to the surroundings. A converging glass lens placed in a liquid of higher index becomes diverging, because $(n-1)$ changes sign.

$P=\frac{1}{f}[\text{dioptre}],P_{total}=P_1+P_2+\dots ,\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1{f}_2}$

Powers of thin lenses in contact simply add, which is why an optician writes a prescription as one number. Setting $d=0$ recovers the contact formula — the check to run if you are unsure of the correction term's sign.

This also resolves the opening question. Every point of the lens refracts light from every object point, so covering half the aperture halves the light reaching each image point. The image dims uniformly; nothing vanishes.

Illustration 4

An object and a screen are 90 cm apart. A converging lens between them gives a sharp image at two positions, 30 cm apart. Find its focal length.

Two positions exist because object and image distances can swap — reversibility again. With separation $D$ and displacement $d$:

$f=\frac{D^2-d^2}{4D}=\frac{8100-900}{360}\Rightarrow f=20\text{cm}$

This displacement method is how focal length is measured in the laboratory, because it needs no knowledge of where the optical centre sits inside the glass.

Aberrations

Chromatic aberration cannot occur in a mirror at all, since reflection is wavelength-independent. That is the single strongest argument for reflecting telescopes.

Illustration 5

A converging lens of $f=15$ cm faces a concave mirror of $f=20$ cm, 50 cm away. An object sits 30 cm in front of the lens. Where is the final image?

Lens first, with $u=-30$ and $f=+15$:

$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{15}-\frac{1}{30}=\frac{1}{30}\Rightarrow v=+30\text{cm}$

That image sits 30 cm to the right of the lens, so it is $50-30=20$ cm in front of the mirror.

Mirror second. Its object sits at exactly the focal length, so the reflected rays leave parallel and no arithmetic is needed.

Lens again. A parallel beam returning through the lens converges at its focal point, putting the final image 15 cm to the left of the lens, and real.

Two things made this tractable: taking one element at a time, and noticing at the mirror that the object had landed on $F$, which skipped the algebra entirely. Check for that coincidence before computing anything in a multi-element problem.

6. Prisms and Dispersion

$\delta =i+e-A,n=\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

Deviation is minimum when the passage is symmetric, $i=e$, with the ray inside parallel to the base. For a thin prism this collapses to $\delta =(n-1)A$, which is what most numerical questions actually need.

Dispersion occurs because $n$ depends on wavelength: violet has the highest index and deviates most, red the least.

$\text{angular dispersion}={\delta }_v-{\delta }_r,\omega =\frac{n_v-n_r}{n_y-1}$

Dispersive power depends on the material alone, not the prism angle — which is exactly what lets two prisms of different glasses deviate without dispersing.

Illustration 6

A crown prism of angle $6^{\circ }$ ($n_y=1.52$, $\omega =0.021$) is combined with a flint prism ($n_y=1.65$, $\omega =0.045$) to give deviation without dispersion. Find the flint angle and the net deviation.

${\delta }_1=(n_1-1)A_1=(0.52)(6)=3.12^{\circ }$

No net dispersion means the two angular dispersions cancel, ${\omega }_1{\delta }_1={\omega }_2{\delta }_2$:

${\delta }_2=\frac{(0.021)(3.12)}{0.045}=1.46^{\circ }\Rightarrow A_2=\frac{1.46}{0.65}=2.24^{\circ }$

${\delta }_{net}=3.12-1.46=1.66^{\circ }$

Deviation survives because the two $\delta$ values differ; dispersion dies because the two $\omega \delta$ products match. Reverse the condition — equal deviations, unequal dispersions — and you get a direct-vision spectroscope instead.

7. Optical Instruments

A microscope needs both focal lengths short; a telescope needs the objective long and the eyepiece short. That one contrast explains the whole difference in design. A telescope objective also needs a large aperture, since it must gather light from a faint source.

Reflecting telescopes win on three counts: no chromatic aberration at all, only one surface to figure, and a mirror can be supported from behind while a lens can only be held at its rim.

Illustration 7

A telescope has $f_o=100$ cm and $f_e=2$ cm. Find its magnifying power and tube length in normal adjustment, and the magnification with the final image at the near point.

$M=\frac{f_o}{f_e}=50,L=f_o+f_e=102\text{cm}$

$M_{near}=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)=50\left(1+\frac{2}{25}\right)=54$

Eight per cent more magnification, at the cost of straining the eye to accommodate. Normal adjustment is preferred for long observing sessions precisely because the relaxed eye can hold it.

8. Wavefronts and Huygens' Principle

A wavefront is a surface of constant phase — spherical near a point source, effectively plane far from it.

Huygens' principle: every point on a wavefront is a source of secondary spherical wavelets, and the new wavefront is their envelope.

Both the law of reflection and Snell's law follow from that one construction, which was the first unified explanation of the two.

The construction also settled a long dispute. For light to bend toward the normal on entering water, it must travel slower there. Newton's corpuscular theory required the opposite. Foucault measured it in 1850 and the wave prediction won.

Illustration 8

Use Huygens' construction to obtain Snell's law, and check the result against the speed comparison it implies.

Let a plane wavefront meet the surface at angle $i$. While one end of it travels the distance $v_{1}t$ still in the first medium, the wavelet launched from the other end has spread only $v_{2}t$ into the second. Both distances are set against the same stretch $L$ of the surface:

$L=\frac{v_{1}t}{\sin i}=\frac{v_{2}t}{\sin r}\Rightarrow \frac{\sin i}{\sin r}=\frac{v_1}{v_2}$

Writing $v=c/n$ converts this at once into $n_1\sin i=n_2\sin r$.

Now the check, which is also the historical one. Light bends toward the normal on entering water, so $\sin r<\sin i$, and the relation then forces $v_2<v_1$: light must travel slower in the denser medium. Newton's corpuscular theory required exactly the opposite, which is what made Foucault's measurement decisive rather than merely confirmatory.

9. Interference and Young's Double Slit

Interference requires coherent sources — a constant phase difference held over time. Two independent bulbs can never interfere, because their phase relationship scrambles billions of times a second and the pattern washes out. This is why Young split one beam instead of using two sources.

$\Delta =d\sin \theta \approx \frac{yd}{D},\text{maxima at }n\lambda ,\text{minima at }\left(n+\frac{1}{2}\right)\lambda$

$\beta =\frac{\lambda D}{d},I=4I_0{\cos }^2\left(\frac{\varphi }{2}\right),\frac{I_{max}}{I_{min}}={\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)}^2$

Fringe width is uniform across the pattern, which distinguishes interference from diffraction at a glance.

Trap. Interference conserves energy exactly. The light missing from dark fringes is precisely the excess in the bright ones, and the average intensity across the screen is what it would have been with no interference at all.

Immersing the apparatus in a liquid shrinks $\beta$ by $n$, because the wavelength shortens while $D$ and $d$ are geometric and do not.

Illustration 9

A thin sheet of index $n$ and thickness $t$ covers one slit. Find the fringe shift, and evaluate it for $t=10\mu$m, $n=1.5$, $\lambda =500$ nm.

Inside the sheet the wave travels $t$ at wavelength $\lambda /n$, so it accumulates the phase of $nt$ in vacuum. The extra path is:

${\Delta }_{extra}=(n-1)t\Rightarrow \text{shift}=\frac{(n-1)t}{\lambda }\text{fringes}$

$=\frac{(0.5)(10\times 10^{-6})}{500\times 10^{-9}}=10\text{fringes}$

The whole pattern slides toward the covered slit, since that path must now be shortened geometrically to compensate. Fringe width is untouched — only the pattern's position moves.

10. Diffraction at a Single Slit

$a\sin \theta =n\lambda (n=1,2,3,\dots )\text{gives MINIMA}$

Trap. This is the chapter's single most confusing point. $\Delta =n\lambda$ gives maxima in the double slit and minima in the single slit. The reason is that a single slit's contributions are spread continuously across its width instead of coming from two discrete points — when the path difference across the whole slit is one wavelength, the contributions pair off and cancel.

Central maximum: angular width $2\lambda /a$, linear width $2\lambda D/a$ — twice every other maximum.

Narrowing the slit widens the pattern. Counterintuitive until you notice it is the wave nature asserting itself as the geometry approaches the wavelength.

Illustration 10

In a double slit with $d=3a$, which interference orders are missing?

A fringe disappears where an interference maximum lands on a diffraction minimum:

$d\sin \theta =m\lambda \text{and}a\sin \theta =p\lambda \Rightarrow \frac{m}{p}=\frac{d}{a}=3$

$m=3,6,9,\dots \text{are missing}$

Missing orders are the clearest evidence that both effects operate at once: the fringes are interference, and the envelope deciding which survive is diffraction.

Illustration 11

A double slit has $d=0.5$ mm with each slit $a=0.1$ mm wide, lit by 600 nm light with the screen 1 m away. Find the fringe width, the width of the central diffraction envelope, and how many bright fringes lie inside it.

$\beta =\frac{\lambda D}{d}=\frac{(600\times 10^{-9})(1)}{0.5\times 10^{-3}}=1.2\text{mm}$

$\text{envelope half-width}=\frac{\lambda D}{a}=6\text{mm}\Rightarrow \text{full width }12\text{mm}$

Naively that gives $12/1.2=10$ fringes. But $d/a=5$, so the orders $m=\pm 5$ land exactly on the envelope's first minima and are missing:

$m=0,\pm 1,\pm 2,\pm 3,\pm 4\Rightarrow 9\text{ bright fringes}$

The spacing comes from $d$ and is perfectly uniform; the heights come from $a$. Interference decides where the fringes are, diffraction decides how bright — and which of them disappear.

11. Polarisation

Only transverse waves can be polarised, so polarisation is direct evidence that light is transverse. Sound, being longitudinal, cannot be polarised at all.

Unpolarised light through one polaroid loses exactly half its intensity — the vibrations are randomly oriented and one component survives.

$I=I_0{\cos }^2\theta (\text{Malus}),\tan {\theta }_p=n(\text{Brewster})$

Two crossed polaroids transmit nothing. Insert a third between them at an intermediate angle and light reappears — a genuinely surprising result and a favourite exam question, because each polaroid rotates the transmitted plane as well as attenuating it.

At the Brewster angle the reflected and refracted rays are exactly perpendicular, which is the neatest way to remember the geometry. Polaroid sunglasses exploit it, blocking the horizontally polarised light reflected off roads and water so that glare vanishes while the rest of the scene does not.

Illustration 12

Light reflects off water ($n=4/3$). Find the Brewster angle, and verify the perpendicularity claim.

$\tan {\theta }_p=\frac{4}{3}\Rightarrow {\theta }_p=53.1^{\circ }$

If reflected and refracted rays are perpendicular, the refraction angle must be $90-53.1=36.9^{\circ }$. Check with Snell:

$\frac{\sin 53.1^{\circ }}{\sin 36.9^{\circ }}=\frac{0.800}{0.600}=1.333=n✓$

The perpendicularity is not a separate fact but the same statement as $\tan {\theta }_p=n$, and either can be used to recover the other.

Summary

• Ray and wave optics are two limits of one theory, separated by whether obstacles are large or comparable to 500 nm.
• Blocking half a lens dims the image; it never removes half of it.
• One sign convention, applied mechanically, solves every mirror and lens problem.
• Mirror: $1/v+1/u=1/f$, $m=-v/u$. Lens: $1/v-1/u=1/f$, $m=v/u$. The differences are deliberate and tested.
• $|m|=k$ usually allows two object positions — one real, one virtual.
• Only an object inside $F$ of a concave mirror gives a virtual erect image.
• Apparent depth is $\text{depth}/n$; a slab shifts by $t(1-1/n)$ without magnifying, and layers add.
• $\sin C=1/n$; TIR is the only lossless reflection, and explains fibres, mirages and diamond.
• The lens maker's formula uses the relative index, so a converging lens turns diverging in a denser liquid.
• Powers add in dioptres; separated lenses take $-d/f_1{f}_2$. Displacement method: $f=(D^2-d^2)/4D$.
• Chromatic aberration cannot occur in a mirror — the case for reflecting telescopes.
• Thin prism: $\delta =(n-1)A$. Dispersive power $\omega$ depends on material alone, which allows achromatic combinations.
• Microscope needs both $f$ short; telescope needs $f_o$ long and $f_e$ short, with $M=f_o/f_e$.
• Huygens: wavelets and their envelope; it predicted light travels slower in water, confirmed in 1850.
• Interference needs coherence — two bulbs never interfere. $\beta =\lambda D/d$, uniform, and shrinks by $n$ in a liquid.
• A sheet over one slit shifts the pattern by $(n-1)t/\lambda$ fringes without changing $\beta$.
• Single slit: $a\sin \theta =n\lambda$ gives minima; central maximum is twice as wide as the rest, and a narrower slit spreads the pattern.
• Missing orders occur when $m/p=d/a$ — interference sets the positions, diffraction sets the envelope.
• One polaroid halves unpolarised light; then $I=I_0{\cos }^2\theta$. A third polaroid between crossed ones restores transmission.
• $\tan {\theta }_p=n$, and at that angle the reflected and refracted rays are perpendicular.