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An object of height $5 cm$ is placed perpendicular to the principal axis of a concave lens of focal length $10 cm .$ If the distance of the object from the optical centre of the lens is $20 cm$ , determine the position, nature and size of the image formed.

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- 6.67 cm,

virtual,

0.33

- 20 cm

, virtual,

1

- 6.67 cm

, real,

0.33

- 20 cm

, real,

0.33

answer is A.

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Detailed Solution

Given: the height of object,

h_{o} = 5 cm

Focal length of the concave lens

(f) = - 10 cm

Object distance,

(u) = - 20 cm

Let image distance

= v

By lens formula,

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

\Rightarrow

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

\Rightarrow

\frac{1}{- 10} + \frac{1}{- 20}

\Rightarrow

\frac{- 2 - 1}{20}

\Rightarrow

- \frac{3}{20}

\Rightarrow

v = - \frac{20}{3} cm

v = - 6.67 cm

Negative sign indicates that the nature of the image is virtual.
Magnification,

m = \frac{v}{u}

= \frac{- 6.67}{- 20}

m = 0.33

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