Q199 / 240 · KCET 2026 · Physics
With reference to the figure shown below, match the following:
List-I List-II (a) Angle of reflection (i) $60^\circ$ (b) Value of $\alpha$ (ii) $120^\circ$ (c) Angle of deviation (iii) $30^\circ$
| List-I | List-II | ||
|---|---|---|---|
| (a) | Angle of reflection | (i) | $60^\circ$ |
| (b) | Value of $\alpha$ | (ii) | $120^\circ$ |
| (c) | Angle of deviation | (iii) | $30^\circ$ |
- Aa-i, b-ii, c-iii
- Ba-ii, b-i, c-iii
- Correct answerCa-iii, b-i, c-ii
- Da-iii, b-ii, c-i
Solution
From the given figure, the angle of incidence $i = 30^\circ$.
The angle of reflection $r = 30^\circ$. This matches (a) with (iii).
Since the normal is perpendicular to the mirror surface, $r + \alpha = 90^\circ$.
$\Rightarrow 30^\circ + \alpha = 90^\circ \Rightarrow \alpha = 60^\circ$. This matches (b) with (i).
The angle of deviation $\delta$ is given by $\delta = 180^\circ - (i + r)$.
$\Rightarrow \delta = 180^\circ - (30^\circ + 30^\circ) = 120^\circ$. This matches (c) with (ii).
Therefore, the correct matching is a-iii, b-i, c-ii.
Answer: a-iii, b-i, c-ii
The angle of reflection $r = 30^\circ$. This matches (a) with (iii).
Since the normal is perpendicular to the mirror surface, $r + \alpha = 90^\circ$.
$\Rightarrow 30^\circ + \alpha = 90^\circ \Rightarrow \alpha = 60^\circ$. This matches (b) with (i).
The angle of deviation $\delta$ is given by $\delta = 180^\circ - (i + r)$.
$\Rightarrow \delta = 180^\circ - (30^\circ + 30^\circ) = 120^\circ$. This matches (c) with (ii).
Therefore, the correct matching is a-iii, b-i, c-ii.
Answer: a-iii, b-i, c-ii
Hence, the correct answer is option: 'C'.