# An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

By BYJU'S Exam Prep

Updated on: September 13th, 2023

Given

Height of object = 5 cm

Position of object u = – 25 cm

Focal length of lens f = 10 cm

We have to find the position of image v, size, and nature of the image.

Let us use the formula

1/v – 1/u = 1/f

Now substitute the values

1/v + 1/25 = 1/10

1/v = 1/10 – 1/25

1/v = (5 – 2)/ 50

1/v = 3/50

v = 50/3 = 16.66 cm

The distance of the image on the opposite side of the lens is 16.66 cm.

We know that

Magnification = v/u

m = 16.66/-25

m = -0.66

m = height of the image/ height of the object

-0.66 = height of the image/ 5 cm

Height of the image = -3.3 cm

A negative sign shows that an inverted image is formed

The position of the image is at 16.66 cm on the opposite side of the lens

The size of the image is -3.33 cm on the opposite side of the lens

The nature of the image will be real and inverted

Therefore, the position is at 16.66 cm on the opposite side of the lens, the size is -3.33 cm on the opposite side of the lens and the nature of the image formed will be real and inverted.

Summary:

## An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size, and nature of the image formed.

An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. The position is at 16.66 cm on the opposite side of the lens, the size is -3.33 cm on the opposite side of the lens and the nature of the image formed will be real and inverted.

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