Question

the focal length of a concave lens is 30 cm what will the position of and size of the image is a 30 cm long object is placed at the focus

Answer 1

i think

question is wrong

Answer 2

Answer:

Correct question

The focal length of a concave lens is 30 cm. An object of 30 cm height is placed at its focus. Calculate the size and position of image.

Explanation:

$Given, \: ƒ \: = - 30cm, \\ \: \: \: \: \: \: \:u = - 30cm, \\ (since \: object \: is \: placed \: at \: focus)$

$height \: of \: object \: (o) = 30cm$

$using \: \: \: \: \: \: \: \frac{1}{ƒ} = \frac{1}{v} - \frac{1}{u}$

$⇒ \: \: \: \: \: \: \: \: \: \frac{1}{v} = \frac{1}{ ƒ} + \frac{1}{u} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{v} = \frac{1}{ - 30} + \frac{1}{u}$

$⇒ \: \: \: \: \: \: \: \: \: \: \frac{1}{v} = \frac{ - 1 - 1}{30} = \frac{ - 2}{30}$

$⇒ \: \: \: \: \: \: \: \: \: \: \: \frac{1}{v} = - \frac{1}{15}$

$⇒ \: \: \: \: \: \: \: \: \: \: \: \: \: v = - 15cm.$

Therefore, image will be forced at 15 cm distance in front of the mirror.

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: m = \frac{I}{O} = \frac{ - v}{u}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{I}{30} = \frac{ - ( - 15)}{30}$

$⇒ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: I = \frac{15 \times 30}{30}$

$⇒ \: \: \: \: \: \: \: \: \: \: \: \: \: \: I = 15cm.$

i.e. Image will be erect and 15cm high.