Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
914 views
in Physics by (29.6k points)
closed by

A 2 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image. Also, find its magnification.

1 Answer

+1 vote
by (29.7k points)
selected by
 
Best answer

Here, u = -15 cm, f = +10 cm 

\(\frac{1}{v}\)\(\frac{1}{u}\) = \(\frac{1}{f}\) ⇒ \(\frac{1}{v}\) = \(\frac{1}{f}\) + \(\frac{1}{u}\) = \(\frac{1}{10}\) + \(\frac{1}{-15}\)  

= \(\frac{1}{30}\)

∴ ν = + 30 cm 

Magnification, m = \(\frac{v}{u}\) = \(\frac{30}{-15}\) = −2

 Nature of image: Real, inverted, magnified as |m| = 2>0 

Size of image(hi): m = \(\frac{h_i}{h_o}\)

∴ hi = m x ho = -2 x 2 = -4 cm

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...