A green light is incident from the water to the air-water interface at the critical angle ($$\theta_c$$). Select the correct statement.

We begin with Snell’s law, which for any interface between two media is written as $$n_1\sin\theta_1 = n_2\sin\theta_2,$$ where $$n_1$$ and $$n_2$$ are the refractive indices of the first and the second medium, and $$\theta_1$$ and $$\theta_2$$ are the corresponding angles measured from the normal.

When light travels from a denser medium to a rarer medium, total internal reflection becomes possible. The critical angle $$\theta_c$$ is that special incident angle in the denser medium for which the refracted ray just grazes the interface, so $$\theta_2 = 90^\circ.$$ Substituting $$\theta_2 = 90^\circ$$ in Snell’s law gives

$$n_{\text{water}}\sin\theta_c = n_{\text{air}}\sin 90^\circ.$$

Because $$\sin 90^\circ = 1,$$ we obtain the well-known formula for the critical angle:

$$\sin\theta_c = \dfrac{n_{\text{air}}}{n_{\text{water}}} = \dfrac{1}{n_{\text{water}}} \quad\text{(since } n_{\text{air}}\approx 1\text{).}$$

Now the refractive index of water is not the same for all wavelengths; it decreases as wavelength increases (normal dispersion). Visible red light (longer wavelength, lower frequency) therefore has a smaller refractive index than green light, while violet light (shorter wavelength, higher frequency) has a larger index than green.

Let us write this relationship symbolically:

$$\lambda_{\text{red}} > \lambda_{\text{green}} > \lambda_{\text{violet}},$$ $$n_{\text{red}} < n_{\text{green}} < n_{\text{violet}}.$$

Using the critical-angle formula $$\sin\theta_c = 1/n_{\text{water}},$$ a smaller value of $$n_{\text{water}}$$ gives a larger value of $$\sin\theta_c$$ and hence a larger critical angle. Therefore

$$\theta_{c,\text{red}} > \theta_{c,\text{green}} > \theta_{c,\text{violet}}.$$

In the present problem a green ray is incident exactly at its own critical angle $$\theta_{c,\text{green}}.$$ We examine what happens to other colours at this same incident angle.

• For any colour with $$\theta_c$$ < $$\theta_{c,\text{green}}$$ (that is, $$\theta_{c,\text{violet}}$$, $$\theta_{c,\text{blue}}$$, etc.), the incident angle is greater than its critical angle, so these rays undergo total internal reflection and remain inside the water.

• For any colour with $$\theta_c$$ > $$\theta_{c,\text{green}}$$ (that is, $$\theta_{c,\text{yellow}}$$, $$\theta_{c,\text{orange}}$$, $$\theta_{c,\text{red}}$$, etc.), the incident angle is less than its critical angle, so these rays are not totally internally reflected; instead they are refracted and emerge into the air.

The colours that satisfy $$\theta_{c,\text{colour}} > \theta_{c,\text{green}}$$ are precisely those whose refractive index is smaller than that of green light, i.e. those whose wavelength is larger and frequency is smaller than that of green. Consequently, the part of the visible spectrum having frequency lower than green will pass into the air, while the higher-frequency part (blue, violet) will stay inside the water.

Therefore, the correct statement is:

“The spectrum of visible light whose frequency is less than that of green light will come out to the air medium.”

Hence, the correct answer is Option B.