An Object Is Placed At A Distance Of 50 Cm From A Concave Lens Of Focal Length 20 Cm. Find The Nature And Position Of The Image.

The focal length of the concave lens is negative.

f = −20 cm

u = −50 cm

We know that :\( \frac{1}{f} = \frac{1}{v} – \frac{1}{u}\) \(\Rightarrow \frac{1}{-20} = \frac{1}{v} – \frac{1}{-50}\)

v = -14.3 cm

The nature of the image is virtual and the position of the image is erect.

Explore more such questions and answers at BYJU’S.

Was this answer helpful?

 
   

3 (2)

(5)
(1)

Choose An Option That Best Describes Your Problem

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

Leave a Comment

Your Mobile number and Email id will not be published.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

*

*

BOOK

Free Class

Ask
Question