Q.A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is

The key idea is that different colours of light travel at different speeds in glass due to dispersion, with refractive index $n$ decreasing as wavelength increases (red has the smallest $n$, violet the largest).

1. For a given slab thickness $t$, the time taken by a colour to travel through the slab is $t / v$, where $v = c/n$ is the speed in glass.
2. So the travel time is $t n / c$ — directly proportional to the refractive index $n$ of that colour.
3. Since red light has the smallest $n$ in glass, it takes the least time to cross the slab. Violet, with the largest $n$, takes the longest.

Why Critical Angle Comparison Works — The Intuition

When white light enters a glass slab at normal incidence, all colours enter without bending. Inside the slab, each colour travels at a different speed because the refractive index $n$ depends on wavelength — this is dispersion. For typical glass, $n$ is highest for violet (shortest wavelength) and lowest for red (longest wavelength).

The speed of light in the medium is $v = c/n$. So red, with the smallest $n$, travels fastest. Over a fixed slab thickness $t$, the time taken is $t/v = n t / c$. Since red has the smallest $n$, it takes the least time and emerges first.

But why bring up the critical angle? Because the critical angle $\theta_c$ is defined by $\sin \theta_c = 1/n$ (for glass-to-air). A larger $n$ means a smaller $\theta_c$. So red, with the smallest $n$, has the largest critical angle. This is a handy mnemonic: the colour that bends least on entering (red) also has the largest critical angle and travels fastest inside the medium. The critical angle comparison is a quick way to rank refractive indices without memorising numbers.

Step-by-Step Reasoning

1. Normal incidence means no refraction at the first surface.

   When light enters at $0^\circ$ to the normal, Snell's law gives $n_{\text{air}} \sin 0^\circ = n_{\text{glass}} \sin r$, so $r = 0^\circ$. All colours go straight in along the same path.

2. Inside the slab, each colour travels the same geometric distance — the slab thickness $d$. But the optical path length is $n d$, and the actual time taken is $t = \frac{n d}{c}$.

3. Refractive index varies with colour.

   For ordinary glass, the dispersion curve is:

   • Red: $n \approx 1.51$ (lowest)
   • Yellow: $n \approx 1.52$
   • Green: $n \approx 1.53$
   • Blue: $n \approx 1.54$
   • Violet: $n \approx 1.55$ (highest)

   So the time order is: red < yellow < green < blue < violet.

4. Critical angle as a ranking tool.

   The critical angle for the glass-air interface is $\theta_c = \sin^{-1}(1/n)$. Since $n$ is smallest for red, $\sin^{-1}(1/n)$ is largest for red. This gives the same ranking: red has the largest critical angle, hence the smallest $n$, hence the fastest speed.

5. The first colour to emerge is the fastest.

   Therefore, red light exits the slab first.

Method: Ranking Colours by Speed Inside a Medium (Dispersion Problems)

This method solves qualitative ranking questions — which colour of light travels fastest, arrives first, bends most, or has the largest critical angle inside a transparent medium.

Steps

Step 1: Recall how refractive index varies with colour

In ordinary dispersive media (glass, water), refractive index increases as wavelength decreases:

Step 2: Convert the index ranking into a speed ranking

Since $v = c/n$, a smaller refractive index means a higher speed inside the medium. This immediately gives the speed order: red fastest, violet slowest.

Step 3: Apply the speed ranking to the specific question being asked

For "who exits/arrives first" over an equal path length inside the medium, the fastest colour (smallest $n$, i.e. red) wins. For "who bends most on entering", the colour with the largest $n$ (violet) deviates most from the incidence direction.

Step 4 (Applying to this problem): Cross-check with critical angle if the question involves TIR

The critical angle $\theta_c=\sin^{-1}(1/n)$ moves the opposite way to $n$: smaller $n$ gives a larger critical angle. So red has the largest critical angle and violet the smallest — a consistent alternate route to the same colour ranking, useful whenever a problem phrases itself in terms of critical angle instead of speed directly.