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Geometrical Optics

Section: Physics

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Question : 16 of 56

Marks: +1, -0

An extended object is placed at point

O, 10\ \mathrm{cm}

in front of a convex lens

L_{1}

and a concave lens

L_{2}

10\ \mathrm{cm}

behind it, as shown in the figure. The radii of curvature of all the curved surfaces in both the lenses are

20\ \mathrm{cm}

. The refractive index of both the lenses is

1.5

. The total magnification of this lens system is

[JEE Adv 2021 P1]

$0.4$
$0.8$
$1.3$
$1.6$

Solution:

Form Lens Maker's formula, focal length of convex lens

(f_{1})

\frac{1}{f_{1}}=(\mu-1)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]

=(1.5-1)\left[\frac{1}{20}-\left(\frac{1}{-20}\right)\right]=\frac{1}{20}

\therefore f_{1}=+20\ \mathrm{cm}

Similarly, focal length of concave lens

(f_{2})

\frac{1}{f_{2}}=(\mu-1)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]

\frac{1}{f_{2}}=(1.5-1)\left[-\frac{1}{20}-\frac{1}{20}\right]=\frac{1}{-20}

\therefore f_{2}=-20\ \mathrm{cm}

Now for lens

L_{1}

\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u}

\Rightarrow \frac{1}{v}=\frac{1}{20}+\frac{1}{-10} \qquad \therefore v=-20\ \mathrm{cm}

\therefore

Magnification

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

Agian for lens

L_{2}

u=-20-10=-30\ \mathrm{cm}, f_{2}=-20\ \mathrm{cm}

\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{-20}+\frac{1}{-30}

\Rightarrow \frac{1}{v}=\frac{-5}{60} \therefore v=-12\ \mathrm{cm}

So, magnification,

M_{2}=\frac{v}{u}=\frac{-12}{-30}=\frac{2}{5}

\therefore

Net magnification,

M=M_{1} M_{2}=2 \times \frac{2}{5}=\frac{4}{5}=0.8

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