(a) According to the question, the focal length of the convex lens, f1 = 30 cm the focal length of the convex lens, f2 = –20 cm separation between two lenses = 8 cm.
if a parallel beam of light is incident from the left on the convex lens, f1 = 30 cm, u1 = –∝
As, 1/v1 - 1/u1 = 1/f1 ⇒ 1/v1 = 1/f1 + 1/u1 = 1/30 + 1/(- ∞) = 1/30
∴ v1 = 30 cm
This image acts as a virtual object for the second lens.
f2 = – 20 cm, u2 = +(30 – 8) = + 22 cm.
Using 1/v2 - 1/u2 = 1/f2
1/v2 = 1/f2 + 1/u2 = 1/-20 + 1/22 = (-11 + 10)/220
⇒ 1/v2 = 1/220 ∴ v2 = - 220 cm
Therefore the parallel beam of light appears to damage from a point (220 – 4) = 216 cm from the centre of the two-lens system.
(ii) Let the parallel beam of light be incident from the left on the concave lens.
Then, f1 = – 20 cm. u1 = – ∞
As, 1/v1 = 1/f1 + 1/u1 = 1/-20 + 1/-∞
⇒ 1/v1 = 1/-20 ∴ v1 = - 20cm
This image acts as real image for the second convex lens:
f2 = + 30cm.
u2 = -(20 + 8) = - 28cm.
∴ 1/v2 = 1/f2 + 1/u2
= 1/30 + 1/-28 = (14 - 15)/420 = - 1/420
⇒ v2 = – 420 cm
The parallel beam of light appears to diverge from a point 420 – 4 = 416 cm on the left of the centre of the two lenses.
The answer depends on which side of the lens, the object is placed. The notion of the effective focal length is not meaningful.
(b) Here object distance, u1 = – 40 cm
f1 = 30cm
As, 1/v1 = 1/f1 + 1/u1 = 1/30 + 1/-40 = (4 - 3)/120
∴ v1 = 120cm
∴ Magnification Produced by the convex lens,
m1 = 120/-40 = -3. m1 = |3|
The image formed by the convex lens acts as virtual object for the concave lens.
u2 = + (120 – 8) = 112 cm, f2 = – 20 cm
As, 1/v2 = 1/f2 = 1/u2 = 1/-20 + 1/112 = (-112 + 20)/(20 x 112) = - 92/(20 x 112)
∴ c2 = (- 112 x 20)/92 = - 24.9cm
∴ Magnification produced by the concave lens,
m2 = v2/u2 = (112 x 20)/(92 x 112) = 20/92
∴ Total magnification produced by the combination
m = m1 x m2 = 3 x 20/92 = 0.652
Size of image = m × size of object = 0.652 × 1.5 = 0.98 cm