Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm each, separated by a distance of 50 cm in air (refractive index = 1) as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect i

Question

Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm each, separated by a distance of 50 cm in air (refractive index = 1) as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image formed by this combination has magnification $M_{1}$ . When the set-up is kept in a medium of refractive index $\frac{7}{6}$ , the magnification becomes $M_{2}$ . The magnitude $| \frac{M_{2}}{M_{1}} |$ is ________. [2015]

Smart Achievers · Accepted Answer

(7)

Applying mirror formula,

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ or $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 10} + \frac{1}{15} = \frac{- 15 + 10}{150} = \frac{- 5}{150} = \frac{- 1}{30}$

$∴$ $v = - 30 cm$

And magnification, $m_{1} = - \frac{v}{u} = - \frac{- 30}{- 15} = - 2$

Now for refraction from lens, $u = - (50 - 30) = - 20 cm$

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f_{l}}$ $\Rightarrow \frac{1}{v} - \frac{1}{- 20} = \frac{1}{10}$

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

Magnification, $m_{2} = \frac{v}{u} = \frac{20}{- 20} = - 1$

Magnification produced by the combination,

$M_{1} = m_{1} \times m_{2} = (- 2) \times (- 1) = 2$

Again, when the system is kept in a medium of refractive index $\frac{7}{6} .$

There is no change for the mirror in this case.

For lens, $\frac{1}{f_{l}'} = (\frac{μ_{l}}{μ_{s}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$\frac{1}{f_{l}'} = (\frac{3 / 2}{7 / 6} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$ $\Rightarrow \frac{1}{f_{l}'} = \frac{2}{7} (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

Also, when the lens was in air,

$\frac{1}{f_{l}} = (1.5 - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = \frac{1}{10}$

$∴$ $(\frac{1}{R_{1}} - \frac{1}{R_{2}}) = \frac{1}{5}$

Using this result in Eqn. (i), we get

$\frac{1}{f_{l}'} = \frac{2}{7} \times \frac{1}{5}$ $∴$ $f_{l}' = \frac{35}{2} cm$

Again using lens formula, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f_{l}'}$

$\frac{1}{v} - \frac{1}{- 20} = \frac{2}{35}$ $\Rightarrow \frac{1}{v} = \frac{2}{35} - \frac{1}{20} = \frac{1}{140}$ $∴$ $v = 140 cm$

Magnification, $m_{2}' = \frac{v}{u} = \frac{140}{- 20} = - 7$

Magnification produced by the combination,

$M_{2} = m_{1} \times m_{2}' = (- 2) \times (- 7) = 14$

$∴$ $| \frac{M_{2}}{M_{1}} |$ $= \frac{14}{2} = 7$

Solution

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