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An object is placed at a distance of $30 cm$ and it is $3 cm$ high placed perpendicular to the principal axes of a concave lens of focal length $15 cm$ . The image is formed at a distance of $10 cm$ . What would be the size and nature of the image formed?

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Virtual and erect,

3 cm

Virtual and erect,

1 cm

Virtual and erect,

2 cm

Virtual and erect,

4 cm

answer is B.

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Detailed Solution

An object is placed at a distance of

30 cm

and it is

3 cm

high placed perpendicular to the principal axes of a concave lens of focal length

15 cm

. The image is formed at a distance of

10 cm

. The image is virtual and erect and the size of the image is

1 cm

.
Sign Conversion of concave lens:
Question Image

Here, Image distance

v = - 10 cm

, focal length

f = - 15 cm

, object distance

u = - 30 cm

m = \frac{v}{u}

, where

m =

the magnification of the image

\Rightarrow Image size = \frac{v}{u} \times object size

\Rightarrow I = - \frac{10}{- 30} \times 3

, where

I =

image size

\Rightarrow Image size = 1 cm

A concave lens will always give a virtual, erect and diminished image, irrespective of the position of the object.

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An object is placed at a distance of 30 cm and it is 3 cm high placed perpendicular to the principal axes of a concave lens of focal length 15 cm. The image is formed at a distance of 10 cm. What would be the size and nature of the image formed?

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An object is placed at a distance of $30 cm$ and it is $3 cm$ high placed perpendicular to the principal axes of a concave lens of focal length $15 cm$ . The image is formed at a distance of $10 cm$ . What would be the size and nature of the image formed?