Chapter 10: Wave Optics

The frequency of a light wave is decided entirely by the source that produces it; it is not a property of the medium through which the light travels.

Reflection and refraction are not processes in which fresh light is created. When the incident wave reaches the boundary, its oscillating electric field drives the charged particles (electrons) of the medium into forced oscillations. A forced oscillator always vibrates at the frequency of the driving force, i.e. at the incident frequency $$\nu$$.

These oscillating charges then act as secondary sources and re-radiate the reflected and refracted waves. Since they oscillate at frequency $$\nu$$, the light they send out — both reflected and refracted — also has frequency $$\nu$$.

Hence the reflected and refracted light have exactly the same frequency as the incident light.

No. A reduction in speed does not mean a reduction in the energy carried by the light.

When light passes from a rarer to a denser medium, its frequency $$\nu$$ stays the same (the frequency is fixed by the source). The speed $$v$$ and the wavelength $$\lambda$$ both decrease, related by $$v = \nu\lambda$$.

The energy carried by a light wave is governed by its intensity, which depends on the square of the amplitude of the wave — not on its speed. Equivalently, in the photon picture each photon carries energy $$E = h\nu$$, which depends only on the frequency. Since $$\nu$$ is unchanged, the energy of each photon is unchanged.

Therefore the reduction in speed has nothing to do with the energy of the wave.

In the wave picture, intensity $$\propto$$ (amplitude)$${}^2$$.

In the photon picture, a beam of light is regarded as a stream of photons, each carrying a fixed energy $$h\nu$$. The intensity of the beam is the amount of light energy crossing unit area (held normal to the beam) per unit time. This equals the energy of one photon multiplied by the number of photons crossing that unit area per unit time.

Since each photon's energy $$h\nu$$ is fixed for monochromatic light, the intensity is determined by the number of photons crossing a unit area normal to the beam per unit time — i.e. by the photon flux. If $$n$$ photons cross unit area per unit time, the intensity is $$I = n\,h\nu$$.

A point source emits light equally in all directions. At any given instant, all the points that the light has reached lie on the surface of a sphere centred on the source — each of these points is at the same distance from the source, and the light has taken the same time to reach them, so they are all in the same phase.

The surface joining points of equal phase is, by definition, the wavefront. Hence the wavefront here is a spherical surface centred on the source, and as the light spreads out this sphere grows in radius.

So the wavefront is a diverging spherical wavefront.

It is a defining property of the focus of a convex lens that rays starting from a point source placed exactly at the focus emerge from the lens parallel to one another (and to the principal axis).

A wavefront is always perpendicular to the rays. Since the emergent rays form a parallel beam, the surface perpendicular to all of them is a flat plane.

Hence the wavefront emerging from the lens is a plane wavefront.

A star, being effectively a point source, emits spherical wavefronts. However, the star is enormously far from the Earth, so by the time a wavefront reaches us its radius is astronomically large.

A small portion of a sphere of very large radius is, for all practical purposes, flat — in the same way that a small patch of the Earth's huge surface looks flat to us.

The Earth intercepts only a tiny portion of this gigantic spherical wavefront, and that portion is essentially flat.

Hence the intercepted wavefront is (very nearly) a plane wavefront.

The refractive index $$n$$ of a medium is defined as the ratio of the speed of light in vacuum, $$c$$, to the speed of light in that medium, $$v$$:

$$n = \dfrac{c}{v} \quad\Rightarrow\quad v = \dfrac{c}{n}$$

Substituting $$c = 3.0\times10^8\,\mathrm{m\,s^{-1}}$$ and $$n = 1.5$$:

$$v = \dfrac{3.0\times10^8}{1.5} = 2.0\times10^8\,\mathrm{m\,s^{-1}}$$

No, the speed of light in glass is not independent of the colour of light.

The refractive index of a transparent medium varies with the wavelength (colour) of light — this wavelength-dependence is precisely what causes dispersion. The single value $$n = 1.5$$ quoted in part (a) is only a representative (average) figure.

For glass the refractive index is greater for violet light than for red light:

$$n_{\text{violet}} > n_{\text{red}}$$

Since the speed of light in glass is $$v = c/n$$, a larger refractive index corresponds to a smaller speed. Therefore violet light travels slower than red light in a glass prism.

(This is why, on passing through a prism, violet light is deviated the most and red light the least.)

Data for the Young’s double–slit set-up:

• Distance between the slits and the screen, $$D = 120 \text{ cm}=1.20\,\text{m}$$
• Separation of the slits, $$d = 0.25 \;\text{mm}=2.5\times10^{-4}\,\text{m}$$
• Wavelength used, $$\lambda = 650\;\text{nm}=6.50\times10^{-7}\,\text{m}$$
• Order of the bright fringe required, $$m = 3$$ (third bright fringe)

The linear distance of the m-th bright fringe from the central maximum is

$$x_m = m\,\dfrac{\lambda D}{d}$$

Substituting the given values,

$$x_3 = 3\,\dfrac{\left(6.50\times10^{-7}\,\text{m}\right)\left(1.20\,\text{m}\right)}{2.5\times10^{-4}\,\text{m}}$$

First evaluate the numerator:

$$3\times6.50\times10^{-7}\times1.20 = 2.34\times10^{-6}\,\text{m}$$

Now divide by the denominator:

$$x_3 = \dfrac{2.34\times10^{-6}}{2.5\times10^{-4}} = 0.00936\,\text{m}$$

$$x_3 \approx 9.4\,\text{mm} = 0.94\,\text{cm}$$

Hence, the third bright fringe for the 650 nm wavelength appears about 9.4 mm from the central maximum.

Note on data: the wording of the exercise does not explicitly state the slit separation $$d$$ and the slit-to-screen distance $$D$$. We use the values quoted in the standard NCERT problem, namely

• Slit separation $$d = 2\,\text{mm}=2\times10^{-3}\,\text{m}$$
• Slit-to-screen distance $$D = 1.2\,\text{m}$$
• Wavelengths $$\lambda_1 = 650\,\text{nm}=6.50\times10^{-7}\,\text{m}$$ and $$\lambda_2 = 520\,\text{nm}=5.20\times10^{-7}\,\text{m}$$

Condition for coincidence. Let the $$m_1$$-th bright fringe of $$\lambda_1$$ coincide with the $$m_2$$-th bright fringe of $$\lambda_2$$. Both must lie at the same distance from the central maximum, i.e.

$$\dfrac{m_1\,\lambda_1 D}{d}=\dfrac{m_2\,\lambda_2 D}{d}\quad\Rightarrow\quad m_1\lambda_1 = m_2\lambda_2$$

Hence

$$\dfrac{m_1}{m_2}=\dfrac{\lambda_2}{\lambda_1}=\dfrac{520}{650}=\dfrac{4}{5}$$

The smallest positive integers satisfying this ratio are $$m_1 = 4$$ and $$m_2 = 5$$. So the very first coincidence (away from the central maximum) occurs where the 4th bright fringe of $$650\,\text{nm}$$ falls on top of the 5th bright fringe of $$520\,\text{nm}$$.

Position of this coincidence. Using the bright-fringe formula with $$m_1 = 4$$ and $$\lambda_1 = 650\,\text{nm}$$:

$$x = m_1\,\dfrac{\lambda_1 D}{d}=4\times\dfrac{(6.50\times10^{-7}\,\text{m})(1.2\,\text{m})}{2\times10^{-3}\,\text{m}}$$

Numerator: $$4\times6.50\times10^{-7}\times1.2 = 3.12\times10^{-6}\,\text{m}$$

Dividing by the denominator:

$$x = \dfrac{3.12\times10^{-6}}{2\times10^{-3}}=1.56\times10^{-3}\,\text{m}$$

$$x \approx 1.56\,\text{mm}$$

Check with the other wavelength: $$m_2\,\dfrac{\lambda_2 D}{d}=5\times\dfrac{(5.20\times10^{-7})(1.2)}{2\times10^{-3}}=1.56\times10^{-3}\,\text{m}$$ — identical, as required.

Hence the least distance from the central maximum at which the bright fringes of the two wavelengths coincide is $$1.56\,\text{mm}$$.