Question

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

​

Answer 1

answer: follow on in brainly

Answer 2

Answer:

• Nature of the image : virtual, erect and diminished.

• The image is formed at a distance of 14.3 cm on the same side of the lens as the object is placed.

Explanation:

Given :

An object is placed at a distance of 50 cm from a concave lens of focal length 20 cm

To find :

the nature and position of the image

Solution :

object distance, u = -50 cm

focal length, f = -20 cm [ concave lens ]

image distance, v = ?

By using lens formula,

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

Substituting the values,

$\sf \dfrac{1}{-20}=\dfrac{1}{v}-\dfrac{1}{-50} \\\\ \sf \dfrac{-1}{20}=\dfrac{1}{v}+\dfrac{1}{50} \\\\ \sf \dfrac{1}{v}=\dfrac{-1}{20}-\dfrac{1}{50} \\\\ \sf \dfrac{1}{v}=\dfrac{-1}{10} \bigg(\dfrac{1}{2}+\dfrac{1}{5} \bigg) \\\\ \sf \dfrac{1}{v}=\dfrac{-1}{10} \bigg(\dfrac{5+2}{10} \bigg) \\\\ \sf \dfrac{1}{v}=\dfrac{-1}{10} \bigg(\dfrac{7}{10} \bigg) \\\\ \sf \dfrac{1}{v}=\dfrac{-7}{100} \\\\ \sf v=\dfrac{-100}{7} \\\\ \longrightarrow \sf v=-14.3 \ cm$

The image is formed at a distance of 14.3 cm on the same side of the lens where object is placed.

Magnification :

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

m = -14.3/-50

m = 14.3/50

m = 0.286

∵ The magnification is positive, the image formed is virtual.

Magnification = size of image/size of object

As we got the magnification value 0.286, the size of image formed is smaller than that of the object.

Hence, the image is diminished.