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 of light be incident on the convex lens first.
`f_(1)=30cm, u_(1)=-oo`
Applying the lens formula, we have
`(1)/(v_(1))-(1)/(u_(1))=(1)/(f_(1))`
`(1)/(v_(1))-(1)/(oo)=(1)/(30)`
`v_(1)=30cm" "(because (1)/(oo)=0)`
The image becomes a virtual object for the second lens. Again
`f_(2)=-20cm`
`u_(2)=+(30-8)=+22cm`.
Therefore, `(1)/(v_(2))=-(1)/(20)+(1)/(22)` which gives `v_(2)=-220cm`.
The parallel incident beam appears to diverge from a point `220-4=216cm` from the centre of the two - lens system.
(ii) Let the parallel beam of light be incident from the left on the concave lens first :
`f_(1)=-20cm, u_(1)=-oo`.
gives `v_(1)=-20cm`. This image becomes a real object for the second lens :
`f_(2)=+30cm`,
`u_(2)=-(20+8)=-28cm`.
Therefore, `(1)/(v_(2))=(1)/(30)-(1)/(28)-(1)/(420)`
`"i.e. "v_(2)=-420 cm`
The parallel incident beam appears to diverge from a point `420 -4=416cm` 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.
(b) Given, `u_(1)=-40cm, f_(1)=30cm, h_(1)=1.5cm`
For a lens, `(1)/(v_(1))-(1)/(u_(1))=(1)/(f_(1))`
`(1)/(v_(1))+(1)/(40)=(1)/(30)`
`"i.e. "v_(1)=120cm`
Magnitude of magnification due to the first (convex) lens,
`m_(1)=|(v_(1))/(u_(1))|=(120)/(40)=3,`
`u_(2)=+(120-8)`
`=+112cm" (object is virtual),"`
`f_(2)=-20cm`
`"Again, "(1)/(v_(2))-(1)/(u_(2))=(1)/(f_(2))`
`(1)/(v_(2))=-(1)/(20)+(1)/(112)`
`"i.e. "(1)/(v_(2))=(112-20)/(-2240)=(92)/(-2240)=(92)/(-2240)cm`
`v_(2)=-24.35cm`
Magnitude of magnification due to the second (concave) lens
`m_(2)=|(v_(2))/(u_(2))|=(1)/(112)((2240)/(92))=0.2174`
Net magnitude of magnification
`m=m_(1)xxm_(2)`
`=3xx0.2174=0.6522`
Size of the image, `h_(2)=mxxh_(1)=0.6522xx1.5`
`=0.98cm`