This is a reference to the question posted by Govind Prajapat Refraction of light through a slab of variable refractive index

A monochromatic light wave is incident normally on a glass slab of thickness 𝑑, as shown in the figure. The refractive index of the slab increases linearly from 𝑛1 to 𝑛2 over the height β„Ž. Which of the following statement(s) is(are) true about the light wave emerging out of the slab? enter image description here

(A) It will deflect up by an angle tan-1[(n22-n12)d/2h]

(B) It will deflect up by an angle tan-1[(n2-n1)d/h]

(C) It will not deflect.

(D) The deflection angle depends only on (𝑛2 βˆ’ 𝑛1 ) and not on the individual values of 𝑛1 and 𝑛2.

In the answer by Stevan V. Saban https://physics.stackexchange.com/q/805443, he has mentioned that the individual rays slow down causing the wavefront to deviate, what is the difference between deviation of the rays and the wavefront?

Furthermore, wouldn't these individual rays deviate as seen when you apply Snell's law at each interface through which the rays pass:

enter image description here

$$\cos(\theta + d\theta)(\mu+d\mu)= cos(\theta)\mu$$

$$(\cos(\theta)\cos(d\theta) - \sin(\theta)\sin(d\theta))(\mu + d\mu) = \cos(\theta)\mu $$

$$\mu\cos(\theta)(1-cos(d\theta)) + \sin(\theta)\sin(d\theta)(\mu + d\mu) - \cos(\theta)\cos(d\theta)d\mu = 0$$

Dividing by $d\theta$ , where $d\theta$ and $d\mu$ tend to 0, we get: $$\mu\sin(\theta) - \cos(\theta)\frac{d\mu}{d\theta} = 0$$

Solving this differential equation we get $\lambda \sec(\theta) = \mu$, where lambda depends on where the ray enters the slab.

Further $\tan(\theta) = \frac{dy}{dx}$

Hence,we get the relation: $$ \lambda \sqrt{(\frac{dy}{dx})^2 + 1} = \mu(y) $$

This gives the trajectory of individual rays which I don't think lead to the same resultant rays exiting out of the slab as obtained by Stevan V. Saban https://physics.stackexchange.com/q/805443,

Why is there a difference? Do the individual rays really deviate in the slab? If not why can we not apply Snell's law in this form in the analysis?


1 Answer 1


In a homogeneous medium a wavefront is at right angles to a ray.
In this problem the medium is inhomogeneous so you should not necessarily expect a ray and wavefront to be at right angles to one another.

A ray is infinitely thin and is along the direction of energy transfer.

A wavefront is a surface (line in two dimensions) representing corresponding points of a wave that vibrate in phase.
So it could be a line along peaks, or a line along troughs, etc.

If an (infinitely thin) ray enters the surface mentioned in the problem along a normal to the surface it will travel undeviated through the block and emerge from the other surface undeviated, the rays are straight lines.

enter image description here

$ACE$ is a wavefront and lets at say at time equal zero $ACE$ are crests.

Each crest travels towards the block and reaches the block after a time $\frac Xc$ where $c$ is the speed of light in air.

The wavefront $ACE$ has moved to the edge of the block with rays and wavefronts perpendicular to one another as there has been no change in the medium.

The time it takes each of the rays to travel through the block is $\frac {d\,n_2}{c},\,\frac {d\,n}{c},,\,\frac {d\,n_1}{c},$ and so the crests emerge out of the block at different times and continue onward into the air still travelling in the same direction.

At a time $t = \frac Xc+\frac {d\,n_2}{c}+\frac {x_2}{c}=\frac Xc+\frac {d\,n}{c}+\frac {x}{c}=\frac Xc+\frac {d\,n_1}{c}+\frac {x_1}{c}$ the crests have travelled distances as shown in the diagram and the line $BDF$ represents a new wavefront (a line along which all the waves are crests).

  • $\begingroup$ "and so the crests emerge out of the block at different times and continue onward into the air still travelling in the same direction." After emerging out of the slab, the rays are no longer perpendicular to the wavefront? i.e. no light energy is received if a solar panel is kept parallel to the new wavefront and towards the direction of travel of these new wavefronts but not intercepting the original rays? Or is the ray always perpendicular to the wavefront? $\endgroup$
    – soccerer
    Commented Apr 9 at 6:53
  • $\begingroup$ Aren't all rays perpendicular to the wavefront? If there was a "component" of the ray parallel to the wavefront, then wouldn't the immediate neighbours be in a different phase contradicting the fact that they are in same phase by definition of a wavefront? $\endgroup$
    – soccerer
    Commented Apr 9 at 17:10

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