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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.
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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संबंधित प्रश्न
A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
- the telescope is in normal adjustment (i.e., when the final image is at infinity)?
- the final image is formed at the least distance of distinct vision (25 cm)?
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| L1 | 3 | 8 |
| L2 | 6 | 1 |
| L3 | 10 | 1 |
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