a) Determine the effective focal length of the combination of the two lenses in Exercise, if they are placed 8.0cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of paralel light is incident? Is the notions of 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 40cm. Determine the magnification produced by the two-lens system, and the size of the image.

Explanation

(a) Here, `f_(1) = 30 cm, f_(2) = - 20 cm, d = 8.0 cm, f = ?`
Let a parallel beam be incident on the convex lens first. If 2nd lens were absent, then
`:. u_(1) = oo` and `f_(1) = 30 cm`
As `(1)/(v_(1)) - (1)/(u_(1)) = (1)/(f_(1)) :. (1)/(v_(1)) - (1)/(oo) = (1)/(30) or v_(1) = 30 cm`
This image would now act as a virtual object for 2nd lens.
`:. u_(2) = + (30 - 8) = + 22 cm, v_(2) = ? f_(2) = - 20 cm`
As `(1)/(v_(2)) = (1)/(f_(2)) + (1)/(u_(2)) :. (1)/(v_(2)) = (1)/(-20) + (1)/(22) = (-11 + 10)/(220) = (-1)/(220)`
`v_(2) = - 220 cm`
`:.` Parallel incident beam would appearto diverge from a point `220 - 4 = 216 cm` from the centre of the two lens system.
(ii) Suppose a parallel beam of light from the left is incident first on the concave lens.
`:. u_(1) = -oo, f_(1) = - 20cm, v_(1) = ?`
As `(1)/(v_(1)) - (1)/(u_(1)) = (1)/(f_(1)) :. (1)/(v_(1)) = (1)/(f_(1)) + (1)/(u_(1)) = (1)/(-20) + (1)/(- oo) = (-1)/(20)`
This image acts as a real object for the 2nd lens
`u_(2) = -(20 + 8) = - 28cm, f_(2) = 30cm, v_(2) = ?`
As `(1)/(v_(2)) - (1)/(u_(2)) = (1)/(f_(2)) :. (1)/(v_(2)) = (1)/(f_(2)) + (1)/(u_(2)) = (1)/(30) - (1)/(28) = (14 - 15)/(420)`
`v_(2) = - 420 cm`
`:.` The parallel beam appears to diverge from a point `420 - 4 = 416 cm`, on the left of the centre of the two lens system.
From the above discussion, we observe that the answer depends on which side of the lens system the parallel beam is incident. Therefore, the notion of effective focal length does not seem to be useful here.
(b) Here, `h_(1) = 1.5 cm, u_(1) = - 40 cm, m = ? h_(2) = ?`
For the 1st lens, `(1)/(v_(1)) - (1)/(u_(1)) = (1)/(f_(1))`
`(1)/(v_(1)) = (1)/(f_(1)) + (1)/(u_(1)) = (1)/(30) - (1)/(40) = (4 - 3)/(120) = (1)/(120)`
`v_(1) = 120 cm`.
Magnitude of magnification produced by first lens, `m_(1) = (v_(1))/(u_(1)) = (120)/(40) = 3`
The image formed by a 1st lens acts as virtual object for the 2nd lens
`:. u_(2) = 120 - 8 = 112 cm, f_(2) = - 20 cm, v_(2) = ?`
As `(1)/(v_(2)) - (1)/(u_(2)) = (1)/(f_(2)) :. (1)/(v_(2)) = (1)/(f_(2)) + (1)/(u_(2)) = (1)/(-20) + (1)/(112) = (-112 + 20)/(112 xx 20) = (- 92)/(112 xx 20)`
`v_(2) = (-112 xx 20)/(92)`
Magnitude of magnification produced by second lens `m_(2) = (v_(2))/(u_(2)) = (112 xx 20)/(92 xx 112) = (20)/(92)`
Net magnification produced by the combination `m = m_(1) xx m_(2) = 3 xx (20)/(92) = (60)/(92) = 0.652`
`:.` Size of image, `h_(2) = mh_(1) = 0.652 xx 1.5 = 0.98 cm`