KCET · Physics · Ray Optics
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) | \(30^\circ\) |
- A a-i, b-ii, c-iii
- B a-ii, b-i, c-iii
- C a-iii, b-i, c-ii
- D a-iii, b-ii, c-i
Answer & Solution
Correct Answer
(C) a-iii, b-i, c-ii
Step-by-step Solution
Detailed explanation
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
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