Question

A convex lens of focal length 15 cm and a concave mirror of focal length 30 cm are kept with their optic axes PQ and RS parallel but separated in vertical direction by 0.6 cm as shown in Fig. The distance between the lens and mirror is 30 cm. An upright object AB of height 1.2 cm is placed on the optics axis PQ of the lens at a distance of 20 cm from the lens. If A'B' is the image after refraction from the lens and reflection from the mirror, find the distance of A'B' from the pole of the mirror and obtain its magnification. Also, locate positions of A' and B' with respect to the optic axis RS

Answer 1

For convex lens using sign convection of coordinate geometry,
$u=+20cm,f=-15cm$
So, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\Rightarrow -\frac{1}{15}=\frac{1}{v_1}-\frac{1}{20}$
$\Rightarrow \frac{1}{v_1}=\frac{1}{20}-\frac{1}{15}=\frac{3-4}{60}\Rightarrow v_1=-60cm$
i.e., image is formed at a distance 60 cm to the left of lens L.
Magnification, $m_1=\frac{v_1}{u_1}=-\frac{60}{20}=-3$
This image is real and inverted. It is intercepted by the mirror.
For concave mirror, $u_2=-60+30=-30cm,f_2=+30cm$
So, $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ gives
$\frac{1}{30}=\frac{1}{v_2}-\frac{1}{30}\Rightarrow \frac{1}{v_2}=\frac{1}{30}+\frac{1}{30}=\frac{2}{30}\Rightarrow v_2=15cm$
Magnification,
$m_2=-\frac{v_2}{u_2}=-\frac{15}{-30}=+\frac{1}{2}$
$\therefore$ Net magnification,
$m=m_1\times m_2=(-3)\times \left(\frac{1}{2}\right)=-1.5$
Size of image A'B'=-1.5\times 1.2 cm=-1.8 cm$
Magnification of mirror is half and image of B formed by convex lens is 0.6 cm above RS, so the length of image will be 1.5 cm below RS.
Thus, B' will be 0.3 cm above RD and A' be 1.5 cm below RS