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.

Given that

The height of object = 5cm

Position of object, u = – 25cm

The focal length of the lens, f = 10 cm

We need to find

The position of the image, v =?

Size of the image

Nature of the image

Formula

We know that

1/v – 1/u = 1/f

Substituting the known values in the above equation we get,

1/v + 1/25 = 1/10

=> 1/v = 1/10 – 1/25

=> 1/v = (5 – 2)/50

Hence, 1/v = 3/50

So, v= 50/3 = 16.66 cm

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

Now, we know that

Magnification = v/u

Hence, m = 16.66/-25 = -0.66

Also, we know that

m= height of image/height of the object

Or, -0.66 = height of image / 5 cm

Hence, height of image = -3.3 cm

The negative sign of the height of the image depicts that an inverted image is formed.

So, the position of image = At 16.66 cm on the opposite side of the lens

Size of image = – 3.3 cm at the opposite side of the lens

Nature of image – Real and inverted

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.

Was this answer helpful?

  
   

4.5 (81)

Upvote (105)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

2 Comments

  1. Very good answer

    1. Thanks a lot 🤗🤗

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class

Ask
Question