How will the image formed when an object is placed 12 cm from a convex lens is focal length is 10 cm?

Given: Focal length (f) = 12 cm,

Object distance (u) = −10 cm

To find: i. Image distance (v)

ii. Nature of image

Formula: `1/"f" = 1/"v" - 1/"u"`

Calculation: From formula,

`1/12 = 1/"v" - 1/(- 10)`

∴ `1/"v" = 1/12 - 1/10`

`= (10 - 12)/120`

`= (- 2)/120`

`= - 1/60`

∴ v = - 60 cm

A negative sign indicates an image is formed on the same side as that of the object.

The image is erect, virtual and magnified.

Page 2

Given: Focal length (f) = 20 cm, object distance (u) = – 10 cm,

height of the object (h1) = 5 cm

To find: Image distance (v), height of the image (h2)

Formulae: 

1. `1/"f" = 1/"v" - 1/"u"`
2. `"h"_2/"h"_1 = "v"/"u"`

Calculation: From formula (i),

`1/20 = 1/"v" - 1/(- 10)`

∴ `1/ "v" = 1/20 - 1/10`

`= (1 - 2)/20`

∴ `1/"v" = - 1/20`

∴ v = - 20 cm

As the image distance is negative, the image formed is virtual and on the same side of lens as that of the object.

From formula (ii),

`"h"_2/5 = (- 20)/- 10`

∴ `"h"_2 = 20/10 xx 5`

∴ h2 = 10 cm

The positive sign indicates that the image formed is erect.