Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • A small telescope has an objective lens of focal length 144cm and an eyepiece of focal length 6.0cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

Q 9.13 A small telescope has an objective lens of focal length 144cm and an eyepiece of focal length 6.0cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

Answers (1)

best_answer

The magnifying power of the telescope is given by 

 m=\frac{f_{objective}}{f_{eyepiece}}

Here, given,

focal length of objective lens =  f_{objective}=   144 cm 

focal length of eyepiece lens = f_{eyepiece}=  6 cm 

 m=\frac{f_{objective}}{f_{eyepiece}}=\frac{144}{6}=24

Hence magnifying power of the telescope is 24.

in the telescope distance between the objective and eyepiece, the lens is given by 

  d= f_{objective}+ f_{eyepiece}

d = 144+6=150

Therefore, the distance between the two lenses is 250 cm.

Posted by

Pankaj Sanodiya

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years
anam-khan

CBSE Class 12 accountancy board exam analysis 2024 out; paper rated easy, balanced

Latest Question