Optics 3 Question 11
11. A small object is placed $50 cm$ to the left of a thin convex lens of focal length $30 cm$. A convex spherical mirror of radius of curvature $100 cm$ is 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 $\theta=30^{\circ}$ to the axis of the lens, as shown in the figure.
(2016 Adv.)
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
(a) $(125 / 3,25 / \sqrt{3})$
(b) $(50-25 \sqrt{3}, 25)$
(c) $(0,0)$
(d) $(25,25 \sqrt{3})$
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Answer:
Correct Answer: 11. (d)
Solution:
For Lens
$$ \begin{aligned} \Rightarrow \quad \frac{1}{v}-\frac{1}{u} & =\frac{1}{f} \Rightarrow \quad v=\frac{u f}{u+f} \\ v & =\frac{(-50)(30)}{-50+30}=75 \end{aligned} $$
For Mirror
$$ \begin{array}{r} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \Rightarrow v=\frac{u f}{u-f} \\ \Rightarrow \quad v=\frac{\frac{25 \sqrt{3}}{2}(50)}{\frac{25 \sqrt{3}}{2}-50}=\frac{-50 \sqrt{3}}{4-\sqrt{3}} \end{array} $$
$$ \begin{aligned} & \Rightarrow \quad m=-\frac{v}{u}=\frac{h _2}{h _1} \Rightarrow h _2=-\frac{\frac{-50 \sqrt{3}}{4-\sqrt{3}}}{\frac{25 \sqrt{3}}{2}} \cdot \frac{25}{2} \\ & \Rightarrow \quad h _2=\frac{+50}{4-\sqrt{3}} \end{aligned} $$
The $x$-coordinate of the images
$=50-v \cos 30+h _2 \cos 60 \approx 25$
The $y$-coordinate of the images
$$ =v \sin 30+h _2 \sin 60 \approx 25 \sqrt{3} $$
