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) (i) Let a parallel beam be the incident from the left on the convex lens first
`f_(1)=30cm and u_(1) =-oo`. , give `v_(1) = + 30cm`. This image becomes a virtual object for the second lens.
`f_(2) = -20 cm, u_(2) = + (30 - 8) cm = + 22 cm` which gives, `v_(2) = - 220 cm`. The parallel incident beam appears to diverge from a point 216 cm from the centre of the two-lens system.
(ii) Let the parallel beam be incident from the left on the concave lens first:`f_(1)=-20cm, u_(1)=-oo, " give "v_(1)=-20cm`. This image becomes a real object for the second lens: `f_(2)=+30cm, u_(2)=-(20+8)cm=-28cm` which gives, `v_(2) = - 420cm`. The parallel incident beam appears to diverge from a point 416 cm on the left of the centre of the two-lens system.
Clearly, the answer depends on which side of the lens system the parallel beam is incident. Further we do not have a simple lens equation true for all u (and v) in terms of a definite constant of the system (the constant being determined by f1 and f2 , and the separation between the lenses). The notion of effective focal length, therefore, does not seem to be meaningful for this system.
(b) `u_(1) = - 40 cm, f_(1) = 30 cm," gives "v_(1) = 120 cm`.
Magnitude of magnification due to the first (convex) lens is 3.
` u_(2) = + (120 – 8) cm = +112 cm" (object virtual), "f_(2) = -20 cm` which gives `v_(2)=-(112xx20)/(92)cm`
Magnitude of magnification due to the second (concave) lens `= 20//92`.
Net magnitude of magnification `= 0.652`
Size of the image = 0.98 cm