Question

An object of length 2cm is placed perpendicularly to the principal axis of a convex lens of focal length 10cm.the distance of object from the lens is 15 ern. find the nature position and size of imag.also find the magnification of the image.​

Answer 1

Given:-

Height of the object = 2 cm

Focal length = +10 cm

Object distance (u) = -15 cm

To find:-

The nature, position and size of the image.

Answer:-

Image distance = +30 cm

Magnification = -2

Height of image = -4 cm

Nature = Real and inverted

Solution:-

Given lens is Convex lens.

In case of lens we use lens formula :-

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

Now, put the given value,

$\dfrac{1}{10} = \dfrac{1}{v} - \dfrac{(-1)}{15}$

$\implies$ $\dfrac{1}{10} = \dfrac{1}{v} + \dfrac{1}{15}$

$\implies$ $\dfrac{1}{10} - \dfrac{1}{15} = \dfrac{1}{v}$

Taking L. C. M of 10 and 15 = 30

$\implies$ $\dfrac{3 - 2}{30} = \dfrac{1}{v}$

$\implies$ $\dfrac {1}{30} = \dfrac{1}{v}$

$\boxed{\sf{ V = +30 cm}}$

Now, The formula for magnification of lens is:-

$\boxed{\sf{ M = \frac {V}{U}}}$

$M = \dfrac {30}{-15}$

$M = -2$

Also,

$\dfrac{h_i}{h_o} = \dfrac{v}{u}$

$\dfrac{h_i}{2} = \dfrac{30}{-15}$

$\dfrac {h_i}{2} = -2$

${h_i} = -4$ Cm

The negative Sign of the height of image show that the image is real and inverted.