A telescope has an objective lens of focal length 200 cm and an eye piece with focal length 2 cm. If this telescope is used to see a 50 m tall building at a distance of 2 km, what is the height of the image of the building formed by the objective lens?

Question

A telescope has an objective lens of focal length 200 cm and an eye piece with focal length 2 cm. If this telescope is used to see a 50 m tall building at a distance of 2 km, what is the height of the image of the building formed by the objective lens? (A) 5 cm (B) 10 cm (C) 1 cm (D) 2 cm

Tardigrade Admin · Accepted Answer

A telescope is an optical instrument used to see distant objects.

Since, convex lens is used, from lens formula we have

\frac{1}{f _{o}} = \frac{1}{v _{o}} - \frac{1}{u _{o}}

where

v_{o}

and

u_{o}

are image and object distance respectively.

∴ \frac{1}{v _{o}} = \frac{1}{f _{o}} + \frac{1}{u _{o}}

Given,

f_{o} = 200 c m

u_{o} = - 2 km = - 2 \times 1 0^{5} c m

O = 50 m = 5 \times 1 0^{3} c m

∴ \frac{1}{v _{o}} = \frac{1}{200} + \frac{1}{- 200 \times 1 0 ^{3}}

= \frac{1 0 ^{3} - 1}{200 \times 1 0 ^{3}}

\Rightarrow v_{o} = \frac{200 \times 1 0 ^{3}}{999} c m

Also magnification

m = ∣ ∣ \frac{v _{o}}{u _{o}} ∣ ∣ = \frac{I}{O}

∴ \frac{200 \times 1 0 ^{3}}{999 \times 200 \times 1 0 ^{3}} = \frac{I}{5 \times 1 0 ^{3}}

\Rightarrow I = \frac{5 \times 1 0 ^{3}}{999} = 5 c m

Q. A telescope has an objective lens of focal length 200cm and an eye piece with focal length 2cm. If this telescope is used to see a 50m tall building at a distance of 2km, what is the height of the image of the building formed by the objective lens?

5 cm

10 cm

1 cm

2 cm

Q. A telescope has an objective lens of focal length $200 c m$ and an eye piece with focal length $2 c m$ . If this telescope is used to see a $50 m$ tall building at a distance of $2 km$ , what is the height of the image of the building formed by the objective lens?