A small object is placed 50 cm to the left of a thin convex lens of focal length 30cm. A convex spherical mirror of radius of curvature 100 cm placed to the right of the lens at a distance of 50 cm. The mirror is tilted such that the axis of the mirror is at an angle $θ = 30^{0}$ to the axis of the lens, as shown in the figure
If the origin of the coordinate system is taken to be at the centre of the lens, the coordinates (in cm) of the point (x, y) at which the image is formed are

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$(0, 0)$

$(125 / 3, 25 / \sqrt{3})$

$(25, 25 \sqrt{3})$

$(50 - 25 \sqrt{3}, 25)$

answer is D.

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Detailed Solution

$F o r l e n s,$

$\begin{array}{l} \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ \Rightarrow \frac{1}{30} = \frac{1}{v} - \frac{1}{- 50} \\ \Rightarrow v = 75 cm \end{array}$

$Distance of this image (I_{1}) from pole of mirror along principal axis = 75 - 50 = 25 cm$

$Mirror formula. if mirror is not rotated$

$\begin{array}{l} \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\ \Rightarrow \frac{1}{50} = \frac{1}{v} + \frac{1}{25} \\ \Rightarrow v = - 50 cm \end{array}$

$The image I would have formed as shown had the mirror been straight.$

$But here the mirror is tilted by 30^{\circ} . So, the image will be tilted by 60^{\circ} and will be formed at A .$

$\begin{array}{l} M A = 50 \cos 60^{\circ} = 25 cm \\ I^{'} A = 50 \sin 60^{\circ} = 25 \sqrt{3} cm \\ Since, origin is the center of lens, so coordinates, \\ x = O M - A M = 50 - 25 = 25 cm \\ y = l^{'} A = 25 \sqrt{3} cm \end{array}$