Question:
(a) Determine the effective focal length of the combination of the two lenses in Exercise 9.10, 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 parallel light is incident? Is the notion 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 objectand the convex lens is 40cm. Determine the magnification produced by the two-lens system, and the size of the image.
Answer:
Focal length of the convex lens (f1) =30 cm
Focal length of the convex lens (f2) =30 cm
Distance between two lens (d) = 20 m
(a) According to lens formula
1/v1 - 1/u1 = 1/f1 where, u1 = Object distance = ∞, v1 = Image distance
1/v1 = 1/f1 + 1/u1 = 1/30 -1/∞ = 1/30
v1 = 30 cm
The image acts as a virtual object for the concave lens
Then, applying les formula,
1/v2 = 1/f2 + 1/u2 where, u2 = Object distance =30-d = 30-8 = 8, v2 = Image distance
or, 1/v2 = 1/22 - 1/20 = 10-11/20 = -1/220
v2 = -220 cm
The parallel incident beam appears to diverge from a point that is = (220-d/2)= (220 - 8/2) = 220-4 = 216 cm from the centre of the combination of the two lenses.
(b)
According to lens formula
1/v2 - 1/u2 = 1/f2 where, u2 = Object distance = - ∞, v2 = Image distance
1/v2 = 1/f2 + 1/u2 = 1/-20 +1/- ∞ = -1/20
v2 = -20 cm
The image will act as a real object for the convex lens
Applying les formula,
1/v1 - 1/u1 = 1/f1
Where, u1 = Object distance = -(20 +d) = -(20+8) = -20 -8 = -28,
v1 = Image distance
1/v2 =1/30 +1/-28 = 14-15/420 = -1/420
v2 = -420 cm
Hence, the parallel incident beam appears to diverge.
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