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
137 views
in Physics by (97.1k points)
closed by
When an object is placed 40cm from a diverging lens, its virtual image is formed 20cm from the lens. The focal length and power of lens are
A. F = -20 cm, P = -5 D
B. F = -40 cm, P = -5 D
C. F = -40 cm, P =-2.5 D
D. F = -20 cm, P = -2.5 D

1 Answer

0 votes
by (97.1k points)
selected by
 
Best answer
Correct Answer - C
We have, `(1)/(f)=(1)/(v)-(1)/(u) rArr (1)/(f)=(1)/(-20)-(-(1)/(40))`
`= (-2+1)/(40)=-(1)/(40)`
or `" " f=-40 cm`
Power of the lens, `P=-(200)/(0.40)=2.5 D`

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

...