An astronomical telescope having an objective of focal length 100cm and an eyepiece of focal length 10cm is used in normal adjustment, calculate the separat...
Question 1 Report
An astronomical telescope having an objective of focal length 100cm and an eyepiece of focal length 10cm is used in normal adjustment, calculate the separation of the lenses
Answer Details
The magnifying power of a telescope in normal adjustment is given by the ratio of the focal length of the objective to that of the eyepiece:
Magnifying power, M = fo/fe
where fo is the focal length of the objective and fe is the focal length of the eyepiece.
The separation of the lenses in a telescope is the sum of the focal lengths of the lenses:
separation of lenses = fo + fe
Given that fo = 100cm and fe = 10cm, we can find the magnifying power and the separation of the lenses:
M = fo/fe = 100cm/10cm = 10
separation of lenses = fo + fe = 100cm + 10cm = 110cm = 1.10m
Therefore, the separation of the lenses is 1.10m. Answer:.
Download The App On Google Playstore
Everything you need to excel in your exams
Personalized AI Learning Chat Assistant
200,000+ Exam Questions Across IGCSE, JAMB, WAEC & NECO
Over 3,900 Lesson Notes
Offline Support - Learn Anytime, Anywhere
Green Bridge Timetable
Literature Summaries & Potential Questions
Track Your Performance & Progress
In-depth Explanations for Comprehensive Learning