In a compound microscope,let u_0 and v_0 be t | vedclass

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(B) Using the lens formula for the objective lens:$\frac{1}{v_0} - \frac{1}{u_0} = \frac{1}{f_0}$Here,$u_0$ is negative (object distance),so let $u_0

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$\frac{f_0}{f_0+u_0}$

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(B) Using the lens formula for the objective lens:$\frac{1}{v_0} - \frac{1}{u_0} = \frac{1}{f_0}$Here,$u_0$ is negative (object distance),so let $u_0 = -|u_0|$.$\frac{1}{v_0} + \frac{1}{|u_0|} = \frac{1}{f_0}$Multiplying the entire equation by $v_0$:$1 + \frac{v_0}{|u_0|} = \frac{v_0}{f_0}$Since magnification $m_0 = \frac{v_0}{u_0}$,and for a real image $m_0$ is negative,we use $m_0 = -\frac{v_0}{|u_0|}$.Rearranging the lens formula: $\frac{v_0}{|u_0|} = \frac{v_0}{f_0} - 1 = \frac{v_0 - f_0}{f_0}$.Alternatively,using the standard magnification formula $m = \frac{f}{f+u}$ where $u$ is the object distance (with sign convention $u $m = \frac{f_0}{f_0 + u_0}$.

In a compound microscope,let $u_0$ and $v_0$ be the object distance and image distance respectively. The objective of focal length $f_0$ magnifies a tiny object into a real,inverted image. The linear magnification of the objective is

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