JEE Advanced
Physics
Ray Optics
2021
JEE Advanced 2021 (Paper 1)
JEE Advanced Physics Question (2021) — Solution
Question
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:
Options
- A. 0 . 4
- B. 0 . 8
- C. 1 . 3
- D. 1 . 6
Step-by-step solution
For convex lens, 1 f = μ 2 μ 1 − 1 1 R 1 − 1 R 2 ⇒ 1 f 1 = 3 2 − 1 1 20 − 1 - 20 ⇒ f 1 = + 20   cm Now, image distance from convex, v convex = u f 1 u + f 1 = − 10 × 20 − 10 + 20 ⇒ v convex = − 20   cm This location of image will be object location for concave lens. Therefore, u concave = − 30   cm For concave lens, 1 f 2 = 3 2 − 1 1 − 20 − 1 20 ⇒ f 2 = − 20   cm Now, distance of image from concave, v concave = − 30 × − 20 − 30 − 20 = - 12   cm Now, m = m convex × m concave m = − 20 − 10 × − 12 − 30 = 0 . 8
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