Where,
= Object distance =∞
v1 = Image distance
The image will act as avirtual object for the concave lens.
Applyinglens formula to the concave lens, we have:
Where,
= Object distance
= (30 − d) = 30 − 8 = 22 cm
= Image distance
The parallel incident beam appears to divergefrom a point that is
from the centre ofthe combination of the two lenses.
(ii) Whenthe parallel beam of light is incident, from the left, on the concave lensfirst:
According to the lensformula, we have:
Where,
= Object distance =−∞
= Image distance
The image will act as areal object for the convex lens.
Applyinglens formula to the convex lens, we have:
Where,
= Object distance
= −(20 + d) = −(20 + 8) = −28 cm
= Image distance
Hence, the parallelincident beam appear to diverge from a point that is (420 − 4) 416 cm from theleft of the centre of the combination of the two lenses.
The answer does depend onthe 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 thiscombination.
(b) Height of the image, h1 = 1.5 cm
Objectdistance from the side of the convex lens, 
According to the lensformula:
Where,
=Image distance
Magnification, 
Hence, the magnificationdue to the convex lens is 3.
The image formed by theconvex lens acts as an object for the concave lens.
According to the lensformula:
Where,
= Object distance
= +(120 − 8) = 112 cm.
= Image distance
Magnification, 
Hence, the magnification due to the concave lensis
The magnification producedby the combination of the two lenses is calculated as:
The magnification of thecombination is given as:
Where,
h1 = Object size = 1.5cm
h2 = Size of the image
Hence,the height of the image is 0.98 cm.