Show that, two thin lenses kept in contact, form an achromatic doublet if they satisfy the condition: `ω/f + (w')/(f') = 0`

where the terms have their usual meaning.

#### Solution

The focal length of a single lens is different for different colors. The image formed by a single lens suffers from chromatic aberration. However, it is possible to combine two lenses of different materials and focal lengths to form an achromatic combination in which f_{r} = f_{v} and image is free from chromatic aberration.

According to the lens maker’s formula

`1/f_v = ( μ_v - 1 )( 1/"R"_1 - 1/"R"_2)`

and `1/f_r = ( μ_r - 1 )( 1/"R"_1 - 1/"R"_2)`

`1/f_v - 1/f_r = ( μ_v - μ_r )( 1/"R"_1 - 1/"R"_2)` ....(i)

If f is the mean focal length of the lens, then

`1/f = ( μ - 1 )( 1/"R"_1 - 1/"R"_2)`

or `1/"R"_1 - 1/"R"_2 = 1/(( μ - 1 )f)`

Putting this value in (i), we get

`1/f_v - 1/f_r = ( μ_v - μ_r )/(( μ - 1)f) = "ω"/f` ...(ii)

Similarly, for the second lens of dispersive power ω' and mean focal length f', we write

`1/(f'_v) - 1/(f'_r) = (ω')/(f') ` ...(iii)

If F_{v} and F_{r} are the focal lengths of the combination for violet and red colours respectively, then

`1/f_v = 1/f_v + 1/(f'_v)` ....(iv)

and `1/F_r = 1/f_r + 1/(f'_r)`

For an achromatic combination

F_{v} = F_{r}`1/f_v + 1/(f'_v) = 1/f_r + 1/(f'_r)`

`( 1/f_u - 1/f_r ) + ( 1/(f'_v) + 1/(f'_r) )` = 0

or `ω/f + (ω')/(f')` = 0 which is the required condition.