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A thin converging lens of focal length f = 1.5 m is placed along y-axis such that its optical centre coincides with the origin. A small light source S is placed at (–2.0 m, 0.1 m). A plane mirror inclined at an angle $θ$ is placed such that y co-ordinate of final image is 0.3m. ( $\tan θ = 0.3$ ).

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The value of d is 4.0m

x co-ordinate of final image is 4.0m

x co-ordinate of final image is 5.0m

The value of d is 5.0m

answer is B, C.

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Detailed Solution

For the lens, u = –2.0 m, f = +1.5 m

$∴ \frac{1}{v} = \frac{1}{- 2 .0} + \frac{1}{1 .5} \Rightarrow v = 6 .0 m$

$m = \frac{6 .0}{- 2 .0} = - 3 .0$

$h_{I} = {mh}_{o} = - 3 \times 0.1 = - 0.3 m$

Therefore, y - coordinate of image formed by lens is –0.3 m

$\frac{0 .3}{NP} = \tan θ = 0 .3$

$∴ NP = MP = 1 .0 m$

$d = 6.0 - 1.0 = 5.0 m$

and, x-coordinate of final image I₂ is, $x = d - 1.0 = 4.0 m$

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