#### Question

An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image

#### Solution

Height of the image, *h*_{1} = 1.5 cm

Object distance from` the side of the convex lens, `u_1 = -40 cm`

`|u_1| = 40 cm`

According to the lens formula:

`1/v_1 - 1/u_1 = 1/f_1`

Where

`v_1` = Image distance

`1/v_1 = 1/30 + 1/(-40) = (4-3)/120 = 1/120`

`:. v_1` = 120 cm

Magnification, `m = v_1/|u_1|`

`= 120/40 = 3`

Hence, the magnification due to the convex lens is 3.

The image formed by the convex lens acts as an object for the concave lens.

According to the lens formula:

`1/v_2 - 1/u_2 = 1/f_2`

Where

`u_2` = Object distance

= +(120 - 8) =112 cm

`v_2` = Image distance

`1/v_2 = 1/-20 + 1/112 = (-112 + 20)/2240 = (-92)/2240`

`:.v_2 = -2240/92 cm`

Magnification, `m' = |v_2/u_2|`

`= 2240/92 xx 1/112 = 20/92`

Hence, the magnification due to the concave lens is `20/92`.

The magnification produced by the combination of the two lenses is calculated as: `m xx m'`

`= 3 xx 20/92 = 60/92 = 0.652`

The magnification of the combination is given as:

`h_2/h_1 = 0.652`

`h_2 = 0.652 xx h_1`

Where

*h*_{1} = Object size = 1.5 cm

*h*_{2} = Size of the image

`:. h_2 = 0.652 xx 1.5 = 0.98 cm`

Hence, the height of the image is 0.98 cm.