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Physics Question on Ray optics and optical instruments

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
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

A

0.4

B

0.8

C

1.3

D

1.6

Answer

0.8

Explanation

Solution

Focal length of convex lens (f1) is given by
1f1=(μ−1)(1R1−1R2)\frac{1}{f_1} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)
=(1.5−1)(120−1−20)(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
1f2=(μ−1)(1R1−1R2)\frac{1}{f_2} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)
=(1.5−1)(−120−120)(1.5 - 1)\left(-\frac{1}{20} - \frac{1}{20}\right)
⇒⇒ 1f2=−120⇒f2=−20 cm\frac{1}{{f_2}} = -\frac{1}{20} \Rightarrow f_2 = -20 \, \text{cm}
Using lens formula for convex lens, we have
1v1−1−10=120\frac{1}{{v_1}} - \frac{1}{-10} = \frac{1}{20}
v1=−20cmv_1=−20 cm
m1=v1u1=−20−10=2m_1 = \frac{{v_1}}{{u_1}} = \frac{{-20}}{{-10}} = 2
Again, applying lens formula for concave lens, we have
1v2−1−30=1−20\frac{1}{{v_2}} - \frac{1}{-30} = \frac{1}{-20}
1v2=−120−130=−360−260\frac{1}{{v_2}} = -\frac{1}{20} - \frac{1}{30} = -\frac{3}{60} - \frac{2}{60}
v2=−12cmv_2=−12 cm
m2=v2u2=−12−30=25m_2 = \frac{{v_2}}{{u_2}} = \frac{{-12}}{{-30}} = 25
Therefore, overall magnification is given by
m=m1×m2=2×25=0.8m = m_1 \times m_2 = 2 \times \frac{2}{5} = 0.8
The Correct answer is option (B): 0.8