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प्रश्न
The focal lengths of the objective and eyepiece of a microscope are 1.25 cm and 5 cm respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment.
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उत्तर
`m_e = D/f_e = 25/5 = 5`
m = m0 × me
`m_0 = m/m_e = -30/5 = -6`
`m_0 = v_0/u_o, -6 = v_0/u_o`
v0 = −6ue
`1/f_e = 1/v_0 - 1/u_0`
`= 1/1.25 = 1/(-6u_0) - 1/u_0 = (-1-6)/(6u_0) = -7/(6u_0)`
`= u_0 = (-7xx1.25)/6`
= −1.46 cm
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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 | |||
Assertion: An astronomical telescope has an objective lens having large focal length.
Reason: Magnifying power of an astronomical telescope varies directly with focal length of the objective lens.
