(a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, 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 is 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.

Focal length of the convex lens, f1 = 30 cm

Focal length of the concave lens, f2 = -20 cm

Distance between the two lenses, d = 8.0 cm

Ans (a).

When the parallel beam of light is incident on the convex lens first:

According to the lens formula, we have:

1/v₁ -1/u₁= 1/f1

Where, u1 = Object distance = ∝  and v₁ = Image distance

1/v₁ -1/∝= 1/30

=> v1 = 30 cm

The image will act as a virtual object for the concave lens.

Applying lens formula to the concave lens, we have:

1/v2 -1/u2 = 1/f2

Where, u2 = Object distance = (30 – d) = 30 – 8 = 22 cm

^v2 = Image distance

1/v2 = 1/22-1/20 = (10-11)/220 = -1/220

Therefore, v2 = -220 cm

The parallel incident beam appears to diverge from a point that is (220-d/2 = 220 — 4) i.e. 216 cm from the centre of the combination of the two lenses.

(ii) When the parallel beam of light is incident, from the left, on the concave lens first:

According to the lens formula,

We have:

1/v2 – 1/u2 = 1/f2

=> 1/v2 = 1/f2 + 1/u2

Where, u₂ = Object distance = – ∝

^v2 = Image distance

1/v2 = 1/(-20 ) + 1/(- ∝)  = -1/20

v2 = -20cm

The image will act as a real object for the convex lens.

Applying lens formula to the convex lens, we have:

1/v1 -1/u1 = 1/f1

Where, u₁ = Object distance = – (20 + d) = – (20 + 8) = -28 cm

v1 = Image distance

1/v1 = 1/30 + 1/(-28) = (14-15 )/420 = -1/420

v2 = -420 cm

Hence, the parallel incident beam appear to diverge from a point that is (420 – 4) 416 cm from the left of the centre of the combination of the two lenses.

The answer does depend on the side of the combination at which the parallel beam of light is incident The notion of effective focal length does not seem to be useful for this combination.

Ans (b).

Height of the image, h1 = 1.5 cm

Object distance from the side of the convex lens, u1 = -40 cm

|u1| = 40 cm

According to the lens formula:

1/v1-1/u1=1/f1

Where,

v1= Image distance

1/v1 = 1/30 + 1/(-40) =( 4-3)/120 =1/120

Therefore ,v1 = 120 cm

Magnification, m = v1/|u1| = 120/40 = 3

Hence, the magnification due to the convex lens is 3.

The image formed by the convex lens acts as an object for the concave lens. According to the lens formula:

1/v2 – 1/u2 = 1/f2

Where, u₂ = Object distance = + (120 – 8) = 112 cm

v₂ = image distance

1/v2 = 1/(-20) + 1/112 = ( -112 + 20 )/2240 = -92/2240

Therefore , v2 = -2240/92 cm

Magnification m’ =  |v2/u2| = (2240/92) x (1/112 ) = 20/92

Hence ,the magnitude due to the concave lens is 20/92

The magnification produced by the combination of the two lenses is calculated as:

m x m’ = 3 x 20 /92 = 60/92 = 0.652

The magnification of the combination is given as :

h2/h1 = 0.652

=> h2 = 0.652 x h1

Where,h1 = Object size = 1.5cm

h2 = size of the image

Therefore, h2 = 0.652 x 1.5 = 0.98 cm

Hence, the height of the image is 0.98 cm