Question

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 $50\, cm$. 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.

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

Answer 1

Answer: (D)

$u =-50\, cm,\, f =+30\, cm$
$\frac{1}{ v }-\frac{1}{ u }=\frac{1}{ f }$
$\Rightarrow v =\frac{ fu }{ u + f }$
$\Rightarrow v =\frac{(30)(-50)}{-50+30}$
$v =+75\, cm$
This image will acts as an object for mirror so the image must lie on line $AB$ (reflected ray from pole) Slope of line $AB$ is
$=\tan (\pi-60)=-\sqrt{3}$
$\frac{\text { check }}{\text { slope }}( A )\left(25, \frac{25}{\sqrt{3}}\right) \&(50,0)$
$m =\frac{25 / \sqrt{3}}{+75}=\frac{1}{3 \sqrt{3}}$
(B) $(50-25 \sqrt{3}, 25)(50,0)$
$m =\frac{25}{-25 \sqrt{3}}=-\frac{1}{\sqrt{3}}$
(C) $(0,0)$ slope zero
(D) $(25,25 \sqrt{3}),(50,0)$
$m = \frac{25 \sqrt{3}}{-25}=-\sqrt{3}$