Question

An object is placed at a distance of 60cm from a concave lens of focal length 30cm.
1) Use lens formula to find the distance of the image from the lens.
2) List four characteristic of the image (nature,position,size,erect/inverted) formed by the lens in this case.
3) Draw ray diagram to justify your answer of part (2).

Answer 1

the solution is shown in the image I have attached. I hope the answer is clear.

Answer 2

Answer: The image is virtual,erect and in size one third of the size of the object.

Explanation:

Given that,

Focal length f = -30 cm

Distance of the object u = -60 cm

(I). Using lens formula

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

$\dfrac{1}{-30}=\dfrac{1}{v}-\dfrac{1}{-60}$

$\dfrac{1}{v}=\dfrac{1}{-20}$

$v = -20 cm$

The Distance of the image is 20 cm from the lens.

(II). The magnification is

$m = \dfrac{v}{u}$

$m = \dfrac{20}{60}$

$m = \dfrac{1}{3}$

Hence, The image is virtual,erect and in size one third of the size of the object.