An extended object is placed at point O, 10 cm in front of a convex lens $$L_1$$ and a concave lens $$L_2$$ is placed 10 cm behind it, as shown in the figure. The radii of curvature of all the curved surfaces in both the lenses are 20 cm. The refractive index of both the lenses is 1.5. The total magnification of this lens system is

The refractive index of each lens with respect to air is $$\mu = 1.5$$ and the radii of curvature of every curved surface are $$20 \text{ cm}$$. For any thin lens in air the lens-maker formula is
$$\frac{1}{f} = (\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

Convex lens $$L_1$$
For a symmetrical bi-convex lens, $$R_1 = +20 \text{ cm}$$, $$R_2 = -20 \text{ cm}$$.
$$\frac{1}{f_1}= (1.5-1)\left(\frac{1}{20}-\frac{1}{-20}\right) = 0.5\left(\frac{1}{20}+\frac{1}{20}\right) = 0.5\left(\frac{1}{10}\right)=0.05$$
$$\Rightarrow f_1 = 20 \text{ cm}$$

Concave lens $$L_2$$
For a symmetrical bi-concave lens, $$R_1 = -20 \text{ cm}$$, $$R_2 = +20 \text{ cm}$$.
$$\frac{1}{f_2}= (1.5-1)\left(\frac{1}{-20}-\frac{1}{20}\right) = 0.5\left(-\frac{1}{20}-\frac{1}{20}\right) = 0.5\left(-\frac{1}{10}\right) = -0.05$$
$$\Rightarrow f_2 = -20 \text{ cm}$$

Image formed by $$L_1$$
Object distance from $$L_1$$: $$u_1 = -10 \text{ cm}$$ (negative because object is to the left).
Lens formula: $$\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1}$$

$$\frac{1}{v_1}-\frac{1}{-10}=\frac{1}{20} \;\;\Longrightarrow\;\; \frac{1}{v_1} = \frac{1}{20}-\frac{1}{10}= -\frac{1}{20} \;\;\Longrightarrow\;\; v_1 = -20 \text{ cm}$$

The image $$I_1$$ is therefore 20 cm to the left of $$L_1$$ (virtual and upright).
Linear magnification of $$L_1$$:
$$m_1 = \frac{v_1}{u_1}= \frac{-20}{-10}= 2$$

Object distance for $$L_2$$
The lenses are 10 cm apart, so the coordinate positions (taking $$L_1$$ at $$x=0$$) are:
Object O at $$x=-10\text{ cm}$$, image $$I_1$$ at $$x=-20\text{ cm}$$, lens $$L_2$$ at $$x=+10\text{ cm}$$.
Hence, for $$L_2$$ the object (which is $$I_1$$) lies 30 cm to its left:
$$u_2 = -30 \text{ cm}$$

Image formed by $$L_2$$
Using the lens formula for the concave lens:
$$\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2}$$

$$\frac{1}{v_2}-\frac{1}{-30}=\frac{1}{-20} \;\;\Longrightarrow\;\; \frac{1}{v_2}= -\frac{1}{20}-\frac{1}{30}= -\frac{5}{60}= -\frac{1}{12} \;\;\Longrightarrow\;\; v_2 = -12 \text{ cm}$$

The final image $$I_2$$ is 12 cm to the left of $$L_2$$ (still on the left side of that lens).
Magnification produced by $$L_2$$:
$$m_2 = \frac{v_2}{u_2}= \frac{-12}{-30}=0.4$$

Total magnification of the lens system
$$m = m_1 \times m_2 = 2 \times 0.4 = 0.8$$

Therefore, the total magnification is 0.8.

Option B which is: 0.8