Question

A small object is placed $50cm$ to the left of a thin convex lens of focal length $30cm$. A convex spherical mirror of radius of curvature $100cm$ is placed to the right of the lens at a distance $50cm$. 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)

$u=-50cm,f=+30cm$
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\Rightarrow v=\frac{fu}{u+f}$
$\Rightarrow v=\frac{(30)(-50)}{-50+30}$
$v=+75cm$
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) $(25,25\sqrt{3}),(50,0)$
$m=\frac{25\sqrt{3}}{-25}=-\sqrt{3}$