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(a) A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the ‘effective focal length’ of the combination of the two lenses, if they are placed 8.0 cm 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 placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.

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(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 

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