A point source of light S, placed at a distance $$60cm$$ in front of the centre of a plane mirror of width $$50cm$$, hangs vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror a distance $$1.2m$$ from it (see$$ in the $$figure). The distance between the extreme points where he can see the image of the light source in the mirror is $$____________cm$$.

Question

Infinity Learn · Accepted Answer

Using properties of similar triangles, $$2560=$$$$tan$$⁡θ$$=y1.2$$⇒$$y=0.5m=50cm$$ Thus, length of field of view $$=y+50+y=150cm$$

Plane mirror

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$Using properties of similar triangles,$
$\begin{array}{l} \frac{25}{60} = \tan θ = \frac{y}{1.2} \\ \Rightarrow y = 0.5 m = 50 cm \\ Thus, length of field of view = y + 50 + y = 150 cm \end{array}$

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Plane mirror

Remember concepts with our Masterclasses.

Using properties of similar triangles, 2560=tan⁡θ=y1.2⇒y=0.5m=50cm Thus, length of field of view =y+50+y=150cm

Ready to Test Your Skills?

free study material

$Using properties of similar triangles,$
$\begin{array}{l} \frac{25}{60} = \tan θ = \frac{y}{1.2} \\ \Rightarrow y = 0.5 m = 50 cm \\ Thus, length of field of view = y + 50 + y = 150 cm \end{array}$