Question

an object placed at the distance of 50 cm from a concave lens of focal length 20 centimetre find the nature and position of the image​

Answer 1

Answer:-

v = -14.28 cm

Given :-

u = -50 cm

f = -20 cm

To find :-

The nature and position of the image.

Solution:-

By using lens formula :-

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

→$\dfrac{-1}{20}= \dfrac{1}{v}-\dfrac{(-1)}{50}$

→$\dfrac{-1}{20} = \dfrac{1}{v}+\dfrac{1}{50}$

→$\dfrac{1}{v}= \dfrac{-1}{20}-\dfrac{1}{50}$

→$\dfrac{1}{v}= \dfrac{-5 -2}{100}$

→$\dfrac{1}{v} = \dfrac{-7}{100}$

→$v = \dfrac{-100}{7}$

→$v = -14.28 cm$

hence,

The position of the image will be -14.28 cm .

The negative sign shows that the image formed is on same side of the concave lens.

Nature :- Virtual and erect image .