Question

a diverging lens has a focal lenght of 38cm, an object is placed 28cm in front of this lens. calculate i) image distance ii) magnification. what is the nature of the image?​

Answer 1

Explanation:

Using the R.I.P. convention u=+38 f=-24 v=???

1/u + 1/v = 1/f

1/v = 1/f - 1/u = -1/24 - 1/38

v = -12.666 — between object and lens

M=v/u = -12.666/38 = -1/3 — erect image, 1/3 size.

Answer 2

Answer:

The correct numerical solution is as follows, taking the focal length of the lens as negative because it is diverging:

1/Di = - 1/24 - 1/38 → Di = - 14.7 cm.

The image is VIRTUAL (Di is negative) diminished as magnification = 14.7/38 = 38.7% of object (14.7 / 38).

Its location is between object and lens.

It is erect, as all virtual images are.