(a) For the telescope described in Exercise 9.34 (a), what is the separation between the objective lens and the eyepiece?

+1 vote
23 views
asked Jan 8, 2018 in Physics by sforrest072 (157,439 points) 60 409 933

(a) For the telescope described in Exercise 9.34 (a), what is the separation between the objective lens and the eyepiece?

 (b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens? 

(c) What is the height of the final image of the tower if it is formed at 25 cm?

1 Answer

+1 vote
answered Jan 8, 2018 by mdsamim (213,225 points) 5 10 15
edited Mar 5, 2018 by Vikash Kumar
 
Best answer

Focal length of the objective lens, fo = 140 cm 

Focal length of the eyepiece, fe = 5 cm

 (a) In normal adjustment, the separation between the objective lens and the

(b) Height of the tower, h1 = 100 m 

Distance of the tower (object) from the telescope, u = 3 km = 3000 m The angle subtended by the tower at the telescope is given as:

The angle subtended by the image produced by the objective lens is given as:

Where, h2 = Height of the image of the tower formed by the objective lens

Therefore, the objective lens forms a 4.7 cm tall image of the tower. 

(c) Image is formed at a distance, d = 25 cm 

The magnification of the eyepiece is given by the relation:

Hence, the height of the final image of the tower is 28.2 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.

One Thought Forever

“There is a close connection between getting up in the world and getting up in the morning.“
– Anon
~~~*****~~~

...