Question

(a) A small object is placed 150 mm away from a diverging lens of focal length 100 mm. (i) Copy the figure below and draw rays to show how an image is formed by the lens. (ii) Calculate the distance of the image from the lens by using the lens formula. (b) The diverging lens in part (a) is replaced by a converging lens also of focal length 100 mm. The object remains in the same position and an image is formed by the converging lens. Compare two properties of this image with those of the image formed by the diverging lens in part (a).

Answer 1

a)
(i)


(ii) u= -150 mm

f = -100 mm

$\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$

$\frac{1}{v}$ - $\frac{1}{-150}$ = $\frac{1}{-100}$

v = - 60 mm

b) u = -150 mm

f = 100 mm

$\frac{1}{v}$ = $\frac{1}{u}$ - $\frac{1}{f}$

$\frac{1}{v}$ - $\frac{1}{-150}$ = $\frac{1}{100}$

$\frac{1}{v}$ = $\frac{1}{100}$ - $\frac{1}{150}$

$\frac{1}{v}$ = $\frac{1}{300}$

v = 300 mm

The image formed by converging lens is real, inverted and magnified (2 times)

It is formed behind the converging lens. On the other hand, image formed by diverging lens is virtual, erect and diminished.

It is formed in front of the diverging lens.