Question

4.An object of 5cm is placed at a
distance of 10cm in front of a bi-
concave lens of focal length 15cm.
Calculate the size and position of the
image and also comment on the
nature of the image.

Urgent pls anyone pls fast ​

Answer 1

Given that,

Size of object, h = 5 cm

Object distance, u = -10 cm

The focal length of a biconcave lens is 15 cm, f = -15 cm

To find,

The size and position of the  image.

Solution,

Lens formula is given by :

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

v is the image distance

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(-15)}+\dfrac{1}{(-10)}\\\\v=-6\ cm$

The image will form at a distance of 6 cm from the lens.

Magnification,

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

h' is the size of the image formed

$h'=\dfrac{vh}{u}\\\\h'=\dfrac{-6\times 5}{-5}\\\\h'=6\ cm$

Hence, the size and position of the  image is 6 cm and 6 cm from the lens respectively. The formed image is virtual and erect.