1.

In an astronomical telescope, focal length of objective lens is `75 cm` and that of eye piece is `5 cm` . Calculate the magnifying power and the distance between the two lenses, when final image of distant object is seen at a distance of `25 cm` from the eye.

Answer» Here, `f_0 = 75 cm, f_e = 5 cm, m = ?`
`L = ? D = 25 cm`
`m = -(f_0)/(f_e)(1 + (f_e)/(d))`
=`-(75)/(5) (1 + (5)/(25)) = -15 xx (6)/(5) = - 18`
From `(1)/(v_e)-(1)/(u_e) = (1)/(f_e)`
`- (1)/(u_e)=(1)/(f_e)-(1)/(v_e)=(1)/(5)-(1)/(-25)=(6)/(25)`
`u_e = (-25)/(6) = -4.17 cm`
Distance between the two lenses, `L = f_0 + |u_e|`
=`75 + 4.17 = 79.17 cm`.


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