Magnifying power of an astronomical telescope is 12 and the image is formed at D.D.V. If the focal length of the objective is 90 cm
Magnifying power of an astronomical telescope is 12 and the image is formed at D.D.V. If the focal length of the objective is 90 cm, what is the focal length of the eyepiece?
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Given: M.P = 12, v = D, f0 = 90 cm,
To find: Focal length of eye piece (fe)
Formula: M.P = \(\frac{f_0}{f_e}\) (1 + \(\frac{f_e}{D}\))
Calculation:
From formula.
12 = \(\frac{90}{f_e}\) (1 + \(\frac{f_e}{25}\))
∴ fe = 10.71 cm
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