Question

Telescope has an objective of focal length 50 cm and an eye-piece of focal length 5 cm. The least distance of distinct vision is 25 cm. The telescope is focused for distinct vision on a scale 2 m away from the objective. Magnification produced is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

2

We use the lens formula and magnification formula for a lens, i.e

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$; $m_o$ = $\frac{v}{u}$

Also, for the eye-piece, $m_e$ $=1+\frac{D}{f}$

Thus, the total magnification comes out to be, M = $m_o\times m_e=\frac{1}{3}\times 6=2$
Magnification produced is 2