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 of $50cm$. The mirror is tilted such that the axis of the mirror is a 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. $(\frac{125}{3},\frac{25}{\sqrt{3}})$
• B. $(25,25\sqrt{3})$
• C. $(50-25\sqrt{3},25)$
• D. $(0,0)$

Answer 1

The correct option is B $(25,25\sqrt{3})$



If $I_1$ is the image formed by convex lens
$u=-50cm,f=+30cm$
Lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v}+\frac{1}{50}=\frac{1}{30}\Rightarrow \frac{1}{v}=\frac{5}{150}-\frac{3}{150}=\frac{2}{150}\Rightarrow v=75cm$

For the mirror, $u=75-50=25cm$
Mirror formula
$\therefore \frac{1}{u}+\frac{1}{v}=\frac{1}{f}\Rightarrow \frac{1}{25}+\frac{1}{v}=\frac{1}{50}\Rightarrow \frac{1}{v}=\frac{1}{50}-\frac{2}{50}$
$\Rightarrow v=-50cm$. i.e. the image will be formed infront of the mirror at $I_2$, if the mirror's axis is same as that of the lens.
But since it is tilted the image will be making an angle of ${60}^{\circ }$ with the original axis at $I_3$


From triangle $I_{3}AP$ , $AP=50cos{60}^{\circ },I_{3}A=50sin{60}^{\circ }$
X- coordinate of final image $x=50-50cos{60}^{\circ }=25cm$
Y- coordinate of final image $y=50sin{60}^{\circ }=25\sqrt{3}cm$