1.

The focal lengths of the objective and eye piece of a microscope are `2 cm and 5 cm` respectively, and the distance between them is `20 cm`. Find the distance of the object from the objective when the final image seen by the eye is `25 cm` from the eye piece. What is the magnifying power ?

Answer» Correct Answer - `u_0 = -2.3 cm, m = 41.5`
Here, `f_0 = 2 cm, f_e = 5 cm, u_0 = ?, m = ?`
As final image is at `25 cm` from eye piece,
therefore, `v_e = -25 cm`
From `(1)/(u_e)=(1)/(v_e)-(1)/(f_e)=(1)/(-25)-(1)/(5)=(-6)/(25)`
`u_e = -(25)/(6) cm`
Distance of image from objective,
`v_0 = 20 - (25)/(6) = (95)/(6) cm`
`(1)/(u_0)=(1)/(v_0)-(1)/(f_0)=(6)/(95) - (1)/(2) = -(83)/(2 xx 95)`
`u_0 = -(2 xx 95)/(83) = -2.3 cm`
Magnifying power `= m_0 xx m_e = (v_0)/(u_0) (1 + (d)/(f_e))`
=`(95//6)/(2 xx 95//83) (1 + (25)/(5)) = 41.5`.


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