A diverging lens has focal length $20 cm$ has an object placed $20 cm$ away from its optical centre. Then the image is formed

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At infinity

At focus

F_{1}

Between focus

F_{1}

and optical centre O

At focus

F_{2}

answer is C.

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Detailed Solution

The image formed Between focus $F_{1}$ and optical centre O.
The diverging lens is also known as the concave lens and it has a negative focal length.
Given,
Focal length, $f = - 20 cm,$
Image distance, $v = - 10 cm$
Object distance is $u$
By using lens maker’s formula, we have
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ Rearranging the equation and putting values,
$\frac{1}{u} = - \frac{1}{10} + \frac{1}{20}$
$\Rightarrow = - 1 / 20$
$u = - 20 cm$
Therefore, when the object is placed at a focal length of $20 cm$ from the optical centre of the divergent lens, then the image is formed in the same side of the lens between Focus and the optical centre. The image formed will be a virtual image, smaller than the object and it is erect.
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