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Question
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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Solution
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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Useful Constants & Relations:
| 1 | Charge of a proton | e | 1.6 × 10-19 C |
| 2 | Speed of light in vacuum | c | 3 × 108 ms-1 |
| 1 u = 931 MeV | |||
