Question 338: Ray Optics

Question

A small telescope has an objective of focal length $$140 \mathrm{~cm}$$ and an eye piece of focal length $$5.0 \mathrm{~cm}$$. The magnifying power of telescope for viewing a distant object is:

Accepted Answer

The magnifying power of a telescope when viewing distant objects can be calculated using the formula:
$$ 	ext{Magnifying Power} (M) = \frac{	ext{Focal Length of Objective Lens} (f_o)}{	ext{Focal Length of Eyepiece Lens} (f_e)} $$
Given in the problem:
Focal Length of Objective Lens, $f_o = 140 \mathrm{~cm}$
Focal Length of Eyepiece Lens, $f_e = 5.0 \mathrm{~cm}$
Substituting these values into the formula gives:
$$ M = \frac{140 \mathrm{~cm}}{5.0 \mathrm{~cm}} = 28 $$
Thus, the magnifying power of the telescope is
28
. The correct answer is
Option B
.

Answer

34

Answer

28

Answer

17

Answer

32

Ray Optics

A small telescope has an objective of focal length $140 \mathrm{~cm} and an eye piece of focal length 5.0 \mathrm{~cm}$. The magnifying power of telescope for viewing a distant object is:

Choose the correct answer: