# A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? - Physics

Numerical

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

#### Solution

Focal length of the objective lens, fo = 144 cm

Focal length of the eyepiece, fe = 6.0 cm

The magnifying power of the telescope is given as:

"m" = ("f"_"o")/"f"_"e"

= 144/6

= 24

The separation between the objective lens and the eyepiece is calculated as:

fo + fe

= 144 + 6

= 150 cm

Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 cm.

Concept: Optical Instruments - Telescope
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 9 Ray Optics and Optical Instruments
Exercise | Q 9.13 | Page 345
NCERT Class 12 Physics Textbook
Chapter 9 Ray Optics and Optical Instruments
Exercise | Q 13 | Page 346
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