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Question

The diagram in Fig. shows an incident ray AO and the reflected ray OB from a plane mirror. The angle AOB Is $$30^0$$. Draw normal on the plane mirror at the point 0 and find : 
1800026_1359365e0b45466499669504ad0f1f63.png


Solution

(i) the angle of incidence 
(ii) the angle of reflection 
ON is normal on the plane mirror at point O
ON is perpendicular on a plane mirror 
Angle of incidence $$\angle i = \angle AON$$ 
and angle of reflection $$\angle r = \angle BON$$ 
Since, $$\angle i - \angle r$$
$$ \angle AOB = 30^0$$
$$\Rightarrow \ \ \angle AON + \angle BON = 30^0$$
$$\Rightarrow \ \ \angle i + \angle i = 30^0$$
$$\Rightarrow \ \  2 \angle i = 30^0$$
$$\Rightarrow \ \ \angle i = 30/2 = 15^0$$
$$\therefore$$ Angle of incidence = $$\angle i = 15^0$$
and Angle of reflection = $$\angle r = 15^0$$ 

1680929_1800026_ans_6fd7cce460374423a18dc144f04043f3.png

Physics
NCERT
Standard XII

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