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
121 views
in Physics by (87.2k points)
closed by
A compound microscope has an objective of focal length `2 cm` and an eye-piece of focal length `3 cm`. An object is placed at a distance of `2.4cm` in front of the field lens. Find the magnifying power of the instrument and length of the tube is (a) final image is at infinity and (b) final image is at least distance of distinct vision.

1 Answer

0 votes
by (84.6k points)
selected by
 
Best answer
`f_(0) = 2cm, f_(e) = 5cm, u_(0) =- 2.4 cm`
(a) `(1)/(v_(0))-(1)/(u_(0)) = (1)/(f_(0))`
`(1)/(v_(0)) - (1)/(-2.4) = (1)/(2)`
`(1)/(v_(0)) = (1)/(2) - (1)/(2.4) = (2.4-2)/(4.8) = (1)/(12) rArr v_(0) = 12 cm`
Tuber length: `L = v_(0) +f_(e) = 12 +5 = 17cm`
`m = (v_(0))/(u_(0)) .(D)/(f_(e)) = (12)/(-2.4) xx (25)/(5) =- 25`
(b) `(1)/(v_(e)) - (1)/(u_(e)) = (1)/(f_(e))`
`(1)/(-25) - (1)/(u_(e)) = (1)/(5) rArr -(1)/(u_(e)) = (1)/(5) +(1)/(25) = (6)/(25)`
`u_(e)=- (25)/(6)cm`
Tube length: `L = v_(0)+|u_(e)| = 12+(25)/(6) = (97)/(6) = 16.2cm`
`m =(v_(0))/(u_(0))(1+(D)/(f_(e))) = (12)/((-2.4)) (1+(25)/(5)) =- 30`

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

...