An extended object is placed at point O, 10 cm in front of a convex lens and a concave lens is placed 10 cm behind it, as shown in the figure. [2021]

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 ________. [2021]

Smart Achievers · Accepted Answer

(2)

From lens maker's formula, focal length of convex lens $(f_{1})$ ,

$\frac{1}{f_{1}} = (μ - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}] = (1.5 - 1) [\frac{1}{20} - (\frac{1}{- 20})]$ $\frac{1}{f_{1}} = \frac{1}{20}$

$∴$ $f_{1} = + 20 cm$

Similarly, focal length of concave lens $(f_{2})$ ,

$\frac{1}{f_{2}} = (μ - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\Rightarrow \frac{1}{f_{2}} = (1.5 - 1) [- \frac{1}{20} - \frac{1}{20}] = - \frac{1}{20}$

$∴$ $f_{2} = - 20 cm$

Now for lens $L_{1}$ ,

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

$\Rightarrow \frac{1}{v} = \frac{1}{20} + \frac{1}{- 10}$ $∴$ $v = - 20 cm$

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

Again for lens $L_{2}$ ,

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

$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{- 20} + \frac{1}{- 30} = \frac{- 5}{60}$

$∴$ $v = - 12 cm$

So, magnification,

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

Hence, net magnification,

$m = m_{1} m_{2} = 2 \times \frac{2}{5} = \frac{4}{5} = 0.8$

Solution

1 0.4

2 0.8

3 1.3

4 1.6

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