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A Small Telescope Has an Objective Lens of Focal Length 150 Cm and an Eye Piece of Focal Length 5 Cm.If this Telescope is Used to View a 100 M High Tower 3 Km Away, - Physics

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प्रश्न

A small telescope has an objective lens of focal length 150 cm and an eye piece of focal length 5 cm. If this telescope is used to view a 100 m high tower 3 km away, find the height of the final image when it is formed 25 cm away from the eye piece.

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उत्तर

The formula for magnifying power is,

Magnifying power, `M = -f_0/f_e  (1+f_e/D)`

where, f0 = Focal length of the objective = 150 cm

fe = Focal length of the eye-piece = 5cm

D = Least distance of distinct vision = 25 cm

`M = -150/5 xx (1+5/25) =-36`

`M =beta/alpha`

`M = tan beta/tan alpha`(As angles α and β are small)

`tan alpha =( \text{Height of object})/(\text { Distance of object from objective}) = H/u = 100/3000 = 1/30`

`M =tan beta/((1/30))`

`tan beta = (-36)/30`

`tan beta = (\text{Height of image})/(\text { Distance of image formation}) = (H')/D`

Thus,

`H' = (-36 xx 25)/30 = -30 cm`

Negative sign indicates that we get an inverted image.

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Optical Instruments - Telescope
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2011-2012 (March) Delhi Set 1
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