Single Choice

A biconvex lens has a radius of curvature of magnitude $$20\ cm$$. Which one of the following options best describe the image formed of an object of height $$2\ cm$$ placed $$30\ cm$$ from the lens?

AVirtual, upright, height $$= 1 cm$$
BVirtual, upright, height $$= 0.5\ cm$$
CReal, inverted, height $$= 4\ cm$$
Correct Answer
DReal, inverted, height $$= 1\ cm$$

Solution

Given,

Radius of curvature, $$R = 20\ cm$$

Height of object, $$h_{0} = 2$$

Object distance, $$u = 30\ cm$$

We have, $$\dfrac {1}{f} = (\mu - 1) \left (\dfrac {1}{R_{1}} - \dfrac {1}{R_{2}}\right )$$

$$= \left (\dfrac {3}{2} - 1\right ) \left [\dfrac {1}{20} - \left (-\dfrac {1}{20}\right )\right ]$$

$$\Rightarrow \dfrac {1}{f} = \left (\dfrac {3}{2} - 1\right ) \times \dfrac {2}{20}$$

$$\therefore f = 20\ cm$$

$$\dfrac {1}{f} = \dfrac {1}{v} - \dfrac {1}{u}$$

$$\Rightarrow \dfrac {1}{20} = \dfrac {1}{v} + \dfrac {1}{30}$$

$$\dfrac {1}{v} = \dfrac {1}{20} -\dfrac {1}{30}$$

$$= \dfrac {10}{600}$$

$$v = 60\ cm$$

$$m = \dfrac {h_{i}}{h_{0}} = \dfrac {v}{u}$$

$$\Rightarrow h_{i} = \dfrac {v}{u} \times h_{0}$$

$$= \dfrac {60}{30}\times 2 = -4\ cm$$

so, image is inverted.

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