1.

The focal lengths of the objective and eye piece of a compound microscope are `4 cm and 6 cm` respectively. If an object is placed at a distance of `6 cm` from the objective, calculate the magnification produced by the microscope. Take distance of distinct vision `= 25 cm`.

Answer» Here, `f_0 = 4 cm, f_e = 6 cm`,
`d = 25 cm, u_0 = -6 cm, m = ?`
If `v_0` is distance of image of object formed by the objective lens, then from
`(1)/(v_0)-(1)/(u_0)=(1)/(f_0)`
`(1)/(v_0)=(1)/(f_0)+(1)/(u_0)=(1)/(4)+(1)/(-6)=(1)/(12)`
`v_0 = 12 cm`
`m = (v_0)/(u_0)(1 + (d)/(f_e)) = (12)/(-6)(1 + (25)/(6)) = -2 xx (31)/(6)`
`m = -10.33`.


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