A microscope has an objective of focal length | vedclass

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(C) For a relaxed eye,the final image is formed at infinity. The distance between the objective and the eyepiece is given by $L = v_0 + f_e$,where $v_

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$140$

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(C) For a relaxed eye,the final image is formed at infinity. The distance between the objective and the eyepiece is given by $L = v_0 + f_e$,where $v_0$ is the image distance from the objective and $f_e$ is the focal length of the eyepiece.Given $L = 25\, cm$ and $f_e = 2.5\, cm$,we have $v_0 = L - f_e = 25 - 2.5 = 22.5\, cm$.Using the lens formula for the objective lens,$\frac{1}{v_0} - \frac{1}{u_0} = \frac{1}{f_0}$,where $f_0 = 1.5\, cm$:$\frac{1}{22.5} - \frac{1}{u_0} = \frac{1}{1.5} \implies \frac{1}{u_0} = \frac{1}{22.5} - \frac{1}{1.5} = \frac{1 - 15}{22.5} = -\frac{14}{22.5}$.Thus,$|u_0| = \frac{22.5}{14} \approx 1.607\, cm$.The magnification for a relaxed eye is given by $m = m_0 	imes m_e = (\frac{v_0}{|u_0|}) 	imes (\frac{D}{f_e})$,where $D = 25\, cm$ is the least distance of distinct vision.$m = (\frac{22.5}{1.607}) 	imes (\frac{25}{2.5}) = 14 	imes 10 = 140$.

$A$ microscope has an objective of focal length $1.5\, cm$ and an eye-piece of focal length $2.5\, cm$. If the distance between objective and eye-piece is $25\, cm$,the approximate value of magnification produced for a relaxed eye is?

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