**(a)** Determine the ‘effective focal length’ of the combination of the two lenses, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of the effective focal length of this system useful at all?

**(b)** 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

Focal length of the convex lens, f_{1} = 30 cm

Focal length of the concave lens, f_{2} = −20 cm

Distance between the two lenses, d = 8.0 cm

**(a) **(i) When the parallel beam of light is incident on the convex lens first:

According to the lens formula, we have:

`1/"v"_1 - 1/"u"_1 = 1/"f"_1`

Where

`"u"_1` = Object distance = ∞

v_{1} = Image distance

`1/"v"_1 = 1/30- 1/∞ = 1/30`

∴ v_{1} = 30 cm

The image will act as a virtual object for the concave lens.

Applying lens formula to the concave lens, we have:

`1/"v"_2 - 1/"u"_2 = 1/"f"_2`

Where,

`"u"_2` = Object distance

= (30 − d) = 30 − 8 = 22 cm

`"v"_2` = Image distance

`1/"v"_2 = 1/22 - 1/20 = (10-11)/220 = (-1)/220`

∴ v_{2} = −220 cm

The parallel incident beam appears to diverge from a point that is `(220 - "d"/2 = 220 - 4)216 "cm"` from the centre of the combination of the two lenses.

(ii) When the parallel beam of light is incident, from the left, on the concave lens first:

According to the lens formula, we have:

`1/"v"_2 - 1/"u"_2 = 1/"f"_2`

`1/"v"_2 = 1/"f"_2 + 1/"u"_2`

Where

`"u"_2` = Object distance = −∞

`"v"_2` = Image distance

`1/"v"_2 = 1/-20 + 1/-∞ = -1/20`

∴ v_{2} = −20 cm

The image will act as a real object for the convex lens.

Applying lens formula to the convex lens, we have:

`1/"v"_1 - 1/"u"_1 = 1/"f"_1`

Where,

`"u"_1` = Object distance

= −(20 + d) = −(20 + 8) = −28 cm

`1/"v"_1 = 1/30 + 1/-28 = (14 - 15)/420 = (-1)/420`

∴ v_{2} = −420 cm

Hence, the parallel incident beam appears to diverge from a point that is (420 − 4) 416 cm from the left of the centre of the combination of the two lenses.

The answer does depend on the side of the combination at which the parallel beam of light is incident. The notion of effective focal length does not seem to be useful for this combination.

**(b)** 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.