A light ray passing from air into water at an angle of 30º from the normal in air would

1. Analyze the question

We are looking at the behavior of a light ray as it travels from a less dense medium (air) into a denser medium (water). We need to determine how the angle of refraction compares to the angle of incidence ($30^\circ$).

2. Explain the concept

This problem is governed by Snell's Law of Refraction, which states:

Where:

• $n_1$ is the refractive index of the first medium (air $\approx 1.0$)
• $n_2$ is the refractive index of the second medium (water $\approx 1.33$)
• $i$ is the angle of incidence ($30^\circ$)
• $r$ is the angle of refraction

The Principle: When light travels from an optically less dense medium (like air) to an optically denser medium (like water), the light ray bends towards the normal. This means the angle of refraction ($r$) will be smaller than the angle of incidence ($i$).

3. Show the solution

• Step 1: Identify the refractive indices. Air has a refractive index of approximately 1.0, while water has a refractive index of approximately 1.33.
• Step 2: Apply the rule of refraction. Since $n_{water} > n_{air}$, the light ray must bend towards the normal.
• Step 3: Compare the angles. Because the ray bends towards the normal, the angle it makes with the normal in the water ($r$) must be smaller than the angle it made with the normal in the air ($i$).
• Step 4: Calculation (conceptual): $$\sin(r) = \frac{n_1 \sin(i)}{n_2} = \frac{1.0 \times \sin(30^\circ)}{1.33}$$Since$\sin(30^\circ) = 0.5$, then$\sin(r) = \frac{0.5}{1.33} \approx 0.376$. Since$\sin(r) < \sin(30^\circ)$, it follows that$r < 30^\circ$.

4. Identify the answer

The angle of refraction is less than the angle of incidence because the light is moving into a denser medium. This matches option D.

5. Tip

Use the mnemonic "DENT": Denser to Entry (or air), Normal Turns away. Conversely, when going from Air to Water (Less to Denser), the light bends towards the normal, making the angle smaller.