Q.You are given four sources of light each one providing a light of a single colour – red, blue, green and yellow. Suppose the angle of refraction for a beam of yellow light corresponding to a particular angle of incidence at the interface of two media is $90^{\circ}$. Which of the following statements is correct if the source of yellow light is replaced with that of other lights without changing the angle of incidence?

The critical angle depends on colour: $n_{red} < n_{yellow} < n_{green} < n_{blue}$, so $C_{red} > C_{yellow} > C_{green} > C_{blue}$ (since $\sin C = n_2/n_1$).

The given incidence angle equals yellow's critical angle, $i = C_{yellow}$. Since $C_{red} > i$, red still refracts (bending away from the normal, as always happens on entering a rarer medium). Since $C_{green} < i$ and $C_{blue} < i$, both green and blue exceed their own critical angles and undergo total internal reflec …

Setting up

The angle of refraction for yellow light is given as $90^\circ$ for a particular angle of incidence $i$. An angle of refraction of $90^\circ$ means the refracted ray grazes the interface — by definition, this happens exactly when the angle of incidence equals the critical angle for that colour:

The same angle of incidence $i$ is kept fixed as the source is swapped to red, blue, and green.

How the critical angle changes with colour

For an ordinary dispersive medium (normal dispersion), the refractive index increases as wavelength decreases:

The critical angle for light going from the denser medium ($n_1$) to the rarer medium ($n_2$) is

Since $n_2$ (the rarer medium) is the same for every colour, a larger $n_1$ gives a smaller $C$. So the critical angles are ordered exactly opposite to the refractive indices:

and $i = C_{yellow}$ sits in the middle of this list.

Checking each option

1. Red — $C_{red} > i$, so the incidence angle $i$ is less than red's own critical angle. Red is refracted, not totally internally reflected. (a) is false.
2. Since red is refracted (going from the denser medium into the rarer medium), it bends away from the normal, not towards it — that is how refraction into a rarer medium always works. (b) is false. …

Method: Predicting TIR Outcomes When Swapping Colours at a Fixed Angle of Incidence

This method solves problems where one colour is stated to refract at exactly $90^\circ$ (i.e. it sits exactly at its own critical angle) at a given, fixed angle of incidence, and you must predict what happens to other colours at that SAME incidence angle.

Steps

Step 1: Translate "refraction angle $=90^\circ$" into a critical-angle statement

If a colour's refracted ray grazes the interface ($r=90^\circ$), the given angle of incidence for that colour IS its critical angle: $i = \theta_{c,\text{that colour}}$.

Step 2: Recall how critical angle depends on colour

Since $\theta_c=\sin^{-1}(n_2/n_1)$ and refractive index increases as wavelength decreases (violet/blue $>$ green $>$ yellow $>$ red), critical angle moves the opposite way: red has the largest critical angle, violet/blue the smallest.

Step 3: Rank every other colour's critical angle relative to the reference colour's

Using the ordering from Step 2, decide for each new colour whether its own critical angle is larger or smaller than the reference colour's (and hence than the fixed incidence angle from Step 1). …