JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Geometrical Optics

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

(A)0.4
(B)0.8✓ Correct
(C)1.3
(D)1.6

Answer: (B)

Solution

For the convex lens:

\frac{1}{f _{1}} = (μ - 1) [\frac{1}{R _{1}} - \frac{1}{R _{2}}] = (1.5 - 1) [\frac{1}{20} - \frac{1}{- 20}] = \frac{1}{20}

, so

f_{1} = + 20

cm.

For lens 1 with u = -10 cm:

\frac{1}{v} - \frac{1}{- 10} = \frac{1}{20}

gives v = -20 cm. Magnification

m_{1} = \frac{v}{u} = \frac{- 20}{- 10} = 2

.

For the concave lens:

\frac{1}{f _{2}} = (1.5 - 1) [- \frac{1}{20} - \frac{1}{20}] = - \frac{1}{20}

, so

f_{2} = - 20

cm.

For lens 2, the image from lens 1 acts as object at distance u = -30 cm:

\frac{1}{v} - \frac{1}{- 30} = \frac{1}{- 20}

gives v = -12 cm. Magnification

m_{2} = \frac{- 12}{- 30} = \frac{2}{5}

.

Net magnification =

m_{1} \times m_{2} = 2 \times \frac{2}{5} = 0.8

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Single correct option

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JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Geometrical Optics

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