Wave Optics: common doubts

Question 1

Which model of light predicted that light travels slower in a denser medium?

Accepted Answer

The wave model did. In the wave picture, light bends towards the normal on entering a denser medium only if it moves more slowly there. The corpuscular model explained the same bending by making light speed up in water. When Foucault measured light to be slower in water than in air, the result supported the wave model.

Question 2

What is a wavefront?

Accepted Answer

A wavefront is a surface on which every point of the wave is oscillating in the same phase. Light from a point source spreads out in spherical wavefronts; far from the source, a small part of such a sphere is nearly flat and is treated as a plane wavefront. Rays are drawn at right angles to the wavefronts and show the direction in which the energy travels.

Question 3

What does Huygens principle say?

Accepted Answer

It says every point on a wavefront acts as a source of secondary wavelets that spread out at the speed of the wave, and the new wavefront a moment later is the surface touching all these wavelets. Repeating the step shows how the wave moves forward. Because the wavelets travel at different speeds in different media, the same idea explains both reflection and refraction.

Question 4

Why is there no backward wave in Huygens construction?

Accepted Answer

Because Huygens simply assumed the secondary wavelets are strongest straight ahead and vanish in the backward direction. Without that assumption, the construction would also produce a wave travelling back towards the source, which is never seen. It was an assumption chosen to match observation, and a more complete wave theory later justified it.

Question 5

Does the frequency of light change when it enters water or glass?

Accepted Answer

No, the frequency stays the same; the speed and wavelength both fall by the factor n. Frequency is set by the source, because the number of wave crests reaching the boundary each second must equal the number leaving it. Since v = νλ and the speed drops to c/n, the wavelength becomes λ/n while the frequency is unchanged.

Question 6

How does the wave theory explain Snell's law?

Accepted Answer

When a plane wavefront meets a boundary at an angle, one end enters the new medium first, and its wavelets move at the new speed while the rest of the wavefront is still in the old medium. In the same time the two ends cover distances v₁τ and v₂τ. The geometry of the two right triangles then gives sin i/sin r = v₁/v₂ = n₂/n₁, which is Snell's law.

Question 7

Does light lose energy when it slows down in glass?

Accepted Answer

No. The slowdown is not an energy loss: the frequency stays the same, so in the photon picture each photon keeps its energy hν, and the light regains the speed c when it comes back out into air or vacuum. A beam does lose some energy at a glass surface, by partial reflection, and a little more by absorption inside, but those are separate effects, not caused by the drop in speed.

Question 8

How does a convex lens turn a plane wavefront into a converging one?

Accepted Answer

Because the middle of the wavefront passes through the thickest glass and is held back the most. The edges, crossing thinner glass, get ahead, so the emerging wavefront is curved inward and becomes spherical, converging to the focus. A thin prism works the same way: one part of the wavefront crosses more glass than the other, so the wavefront comes out tilted and the beam turns.

Question 9

Why do all rays from a point on an object take the same time to reach its image?

Accepted Answer

Because the image is where wavefronts from that point converge, and a wavefront joins points of equal phase, which means equal travel time. A ray through the thick middle of a lens covers a short path in slow glass, while a ray near the edge covers a longer path, mostly in faster air. The times balance exactly, so the light arrives in step and forms a sharp image.

Question 10

What is the difference between coherent and incoherent sources?

Accepted Answer

Coherent sources keep a constant phase difference between them, while the phase difference between incoherent sources changes randomly and rapidly. With coherent sources the intensity at each point stays fixed, so a steady interference pattern forms. With incoherent sources any pattern shifts so fast that it averages out, the intensities simply add, and no fringes are seen.

Question 11

Why is the intensity at a bright fringe 4I₀ and not 2I₀?

Accepted Answer

Because at a bright fringe the amplitudes add, not the intensities. Two equal waves of amplitude a meeting in phase give amplitude 2a, and intensity goes as amplitude squared, so it is four times that of one wave. Energy is still conserved: the dark fringes get nothing, and averaged over the pattern the intensity is 2I₀. The formula I = 4I₀ cos²(φ/2) captures both.

Question 12

Why don't two separate lamps shining on two slits produce interference fringes?

Accepted Answer

Because independent sources are incoherent. Light from an ordinary source suffers sudden phase jumps about every 10⁻¹⁰ s, and two separate lamps jump independently, so their phase difference keeps changing. Any fringes vanish far too quickly to be seen, and the screen shows uniform intensity. Young solved this by lighting both slits from one pinhole, so every phase jump reaches the two slits together.

Question 13

What does fringe width depend on in Young's double-slit experiment?

Accepted Answer

Fringe width is β = Dλ/d, so it increases with the screen distance D and the wavelength λ, and decreases as the slit separation d increases. Red light therefore gives wider fringes than blue light. If the whole apparatus is placed in a liquid of refractive index n, the wavelength becomes λ/n, and the fringes become narrower by the same factor.

Question 14

Why is the central fringe in Young's experiment bright?

Accepted Answer

Because every point on the central line is equally far from the two slits, so the path difference is zero and the waves arrive in phase. Since both slits are lit by the same source, they start in step, and with no extra path they stay in step, so their amplitudes add. This is the n = 0 bright fringe, with the others equally spaced on either side at intervals of Dλ/d.

Question 15

What is the condition for dark bands in single-slit diffraction?

Accepted Answer

Dark bands occur where a sin θ ≈ nλ, with n = ±1, ±2, …, which looks like the double-slit condition for bright fringes but means the opposite. At these angles the slit can be split into pairs of strips whose wavelets cancel. The secondary maxima lie roughly midway, near (n + ½)λ/a, each fainter than the last, and the central maximum is twice as wide as the others.

Question 16

Why does a narrower slit give a wider diffraction pattern?

Accepted Answer

Because the first dark band lies at θ ≈ λ/a, so making the slit width a smaller pushes it to a larger angle. The central maximum stretches from −λ/a to +λ/a, an angular width of 2λ/a, so halving the slit width doubles the spread. A longer wavelength widens the pattern too. That is why diffraction of light becomes obvious only with very narrow openings.

Question 17

What is the difference between interference and diffraction?

Accepted Answer

There is no fundamental physical difference: both come from adding waves with their phases. By custom, the result of a few sources, such as two slits, is called interference, and that of very many sources, such as the countless wavelets across one slit, is called diffraction. In Young's experiment the observed pattern is really two-slit interference fringes sitting inside each slit's diffraction pattern.

Question 18

Why can light be polarised but sound in air cannot?

Accepted Answer

Because polarisation needs a transverse wave, whose vibration is at right angles to its direction of travel and so can point in different directions. Light is transverse: its electric field oscillates across the beam. Sound in air is longitudinal, vibrating along its direction of travel, so there is only one possible direction and nothing for a polariser to select. Polarisation shows that light is transverse.

Question 19

What is unpolarised light, and how does a polaroid polarise it?

Accepted Answer

Unpolarised light has its electric field changing direction randomly and rapidly, taking every orientation in the plane perpendicular to the beam, as in sunlight or lamplight. A polaroid contains long-chain molecules aligned in one direction; it absorbs the field component along them and passes the component along its pass-axis. Light leaving it vibrates in that single direction, so it is linearly polarised.

Question 20

Why does the first polaroid pass exactly half of unpolarised light?

Accepted Answer

Because unpolarised light has its field spread equally over all directions, and the average of cos²θ over every angle is one half. So whatever the orientation of the first polaroid, it transmits half the incident intensity. Malus' law, I = I₀ cos²θ, applies only from the second polaroid onward, with I₀ the polarised intensity leaving the first. Applying cos²θ to unpolarised light is a common mistake.

Question 21

Why is Malus' law cos²θ and not cos θ?

Accepted Answer

Because a polaroid passes the component of the electric field along its axis, which is E cos θ, while intensity is proportional to the square of the field. Squaring gives I = I₀ cos²θ. So at θ = 60° only a quarter of the polarised light gets through, not half. At θ = 90°, crossed polaroids, the transmitted intensity is zero.

Question 22

How can light get through two crossed polaroids when a third is placed between them?

Accepted Answer

The middle polaroid turns the direction of polarisation partway, so the light reaching the last polaroid is no longer at right angles to it. With the middle one at angle θ to the first, the light emerging from the third has intensity (I₀/4) sin²2θ, where I₀ is the intensity after the first polaroid. This is largest, I₀/4, at θ = 45°, and zero when the middle one lines up with either of the others.

Question 23

When can light be treated as rays, and when must it be treated as a wave?

Accepted Answer

Rays work well whenever the openings and obstacles light meets are much larger than its wavelength, as with ordinary lenses, mirrors and windows. The wave picture is needed when openings are comparable to the wavelength, or when light from coherent sources overlaps: then diffraction and interference appear, and polarisation too, none of which rays can describe. Visible light has λ of only about 400 to 700 nm, so rays suffice in most everyday situations.

Wave Optics: common doubts, answered

About the chapter

When can light be treated as rays, and when must it be treated as a wave?

Light as a wave