Question

(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.

Answer 1

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 

(a) When the parallel beam of light is incident on the convex lens first: According to the lens formula, we have:

The image will act as a virtual object for the concave lens. Applying lens formula to the concave lens, we have:

The parallel incident beam appears to diverge from a point that

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

Applying lens formula to the convex lens, we have:

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. 

(b) Height of the image, h₁ = 1.5 cm

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:

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

The magnification of the combination is given as:

Hence, the height of the image is 0.98 cm.