Question

List four characteristics of the image formed by a concave lens of focal length 20cm when the object is placed at a distance of 40cm from it optical center of the lens

Answer 1

image will be real and inverted

of same size

magnification 1

position of object = position of image

Answer 2

Answer:

The distance of the image is 13.3 cm. The image is virtual and erect and the size of the image is one by third of the size of the object. The magnification of the lens is 0.333.

Explanation:

Given that,

Focal length f = 20 cm

The object distance u = -40 cm

Using lenses formula,

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

$\dfrac{1}{-20}=\dfrac{1}{v}-\dfrac{1}{-40}$

$\dfrac{1}{v}=\dfrac{-3}{40}$

$v = \dfrac{-40}{3}$

$v = -13.3\ cm$

The distance of the image is 13.3 cm from the lens .

The magnification is

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

$m = \dfrac{-13.3}{-40}$

$m=0.333$

The image is virtual and erect. The size of the image is $\dfrac{1}{3}$ of the size of object.

Hence, The distance of the image is 13.3 cm. The image is virtual and erect and the size of the image is one by third of the size of the object. The magnification of the lens is 0.333.