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JEE Advanced 2021 · Paper 1, Physics question 3

An extended object is placed at point O, 10 cm in front of a convex lens L1L_1 and a concave lens L2L_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

Principal axis with an upright object arrow at O, a biconvex lens L1 10 cm to its right, and a biconcave lens L2 a further 10 cm to the right.
Choose one option
Answer and worked solution

Official answer: B

Focal lengths (lens maker's formula, μ=1.5\mu=1.5, ∣R∣=20|R|=20 cm):

  • L1L_1: f=+20f=+20 cm.
  • L2L_2: f=−20f=-20 cm.

First lens. With u=−10u=-10 cm,
1v=120−110  ⟹  v=−20 cm,m1=vu=2.\frac1v=\frac1{20}-\frac1{10}\implies v=-20\text{ cm},\qquad m_1=\frac vu=2.
The image is virtual, 20 cm in front of L1L_1.

Second lens. That image is 30 cm in front of L2L_2, so u=−30u=-30 cm:
1v=−120−130  ⟹  v=−12 cm,m2=−12−30=0.4.\frac1v=-\frac1{20}-\frac1{30}\implies v=-12\text{ cm},\qquad m_2=\frac{-12}{-30}=0.4.

Total magnification. m=m1m2=0.8m=m_1m_2=0.8. (B)

Topics:Combination of lenses and mirrorsThin lenses and lens maker's formula
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