Question

An extended object is placed at point O, 10 cm in front of a convex lens L1 and a concave lens L2 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

• A. 0.4
• B. 0.8
• C. 1.3
• D. 1.6

Answer 1

The Correct Option is B

Focal length of convex lens (f₁) is given by
\(\frac{1}{f_1} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\)
=\((1.5 - 1)\left(\frac{1}{20} - \frac{1}{-20}\right)\)
\(⇒ \)\(\frac{1}{{f_1}} = \frac{1}{20} \Rightarrow f_1 = +20 \, \text{cm}\)
Similarly, focal length of concave lens (f₂) is given as
\(\frac{1}{f_2} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\)
=\((1.5 - 1)\left(-\frac{1}{20} - \frac{1}{20}\right)\)
\(⇒\) \(\frac{1}{{f_2}} = -\frac{1}{20} \Rightarrow f_2 = -20 \, \text{cm}\)
Using lens formula for convex lens, we have
\(\frac{1}{{v_1}} - \frac{1}{-10} = \frac{1}{20}\)
\(v_1=−20 cm\)
\(m_1 = \frac{{v_1}}{{u_1}} = \frac{{-20}}{{-10}} = 2\)
Again, applying lens formula for concave lens, we have
\(\frac{1}{{v_2}} - \frac{1}{-30} = \frac{1}{-20}\)
\(\frac{1}{{v_2}} = -\frac{1}{20} - \frac{1}{30} = -\frac{3}{60} - \frac{2}{60}\)
\(v_2=−12 cm\)
\(m_2 = \frac{{v_2}}{{u_2}} = \frac{{-12}}{{-30}} = 25\)
Therefore, overall magnification is given by
\(m = m_1 \times m_2 = 2 \times \frac{2}{5} = 0.8\)
The Correct answer is option (B): 0.8