A thin converging lens of focal length $f = 1.5 m$ is placed along $y - a x i s$ such that its optical centre coincides with the origin. A small light source $S$ is placed at $(- 2, 0.1) m$ . A plane mirror inclined at an angle $θ$ (where $\tan (θ) = 0.3$ ) is placed as shown in figure, such that $y$ co-ordinate of final image is $0.3$ . Find the distance d, in $m e t r e$ . Also find $x$ co-ordinate of final image, in $m e t r e$ .

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answer is 4.

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Detailed Solution

For the lens, $u = - 2 m, f = \pm 1.5 m$
$\Rightarrow \frac{1}{v} - \frac{1}{- 2} = \frac{1}{1.5} \Rightarrow v = 6 m$
Since, $m = \frac{6}{(- 2)} = - 3$

Therefore, $y$ $c o - o r d i n a t e$ of image formed by the lens is given by
$y = m (0.1) = - 3.0 m$
In triangle $P N I_{2}$ , we have
$\tan (θ) = \frac{I_{2} N}{N P} \Rightarrow \tan (θ) = \frac{0.3}{N P} = 0.3 \{∵ \tan (θ) = 0.3\} \Rightarrow N P = M P = 1 m \Rightarrow d = 6 - 1 = 5 m$
and $x c o - o r d i n a t e$ of final image $I_{2}$ is,
$x = d - 1 = 4 m$