Sarthaks Test
0 votes
6 views
ago in Ray Optics and Optical Instruments by (10.8k points)
closed ago by

A compound microscope consists of an objective lens of focal length 2 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at 

1. Least distance of distinct vision.

2. infinity

1 Answer

+1 vote
ago by (10.9k points)
selected ago by
 
Best answer

1. ve = -25 cm

\(\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}\) = \(\frac{1}{-25}-\frac{1}{6.25}\)

∴ u0 = 5 cm 

Length of the tube, L= |v0| + |ue|

∴ v0 = 15 – 5 = 10

\(\frac{1}{u_0}=\frac{1}{v_0}-\frac{1}{f_0}\) = \(\frac{1}{10}-\frac{1}{2}\)

ue = -2.5 cm.

2.  ∴ v0 = 15 – f= 15 – 6.25 = 8.75

\(\frac{1}{u_0}=\frac{1}{v_0}-\frac{1}{f_0}\)\(\frac{1}{8.75}\) - \(\frac{1}{2}\)

u0 = -2.59 cm

Related questions

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.

...