In a compound microscope,the focal lengths of | vedclass

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(D) For the objective lens: $f_o = 1.5 \, cm$,$u_o = -2 \, cm$. Using the lens formula $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$:$\frac{1}{v_o}

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$11$

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(D) For the objective lens: $f_o = 1.5 \, cm$,$u_o = -2 \, cm$. Using the lens formula $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$:$\frac{1}{v_o} - \frac{1}{-2} = \frac{1}{1.5} \implies \frac{1}{v_o} = \frac{2}{3} - \frac{1}{2} = \frac{4-3}{6} = \frac{1}{6}$.Thus,$v_o = 6 \, cm$.For the eye lens: $f_e = 6.25 \, cm$,$v_e = -25 \, cm$. Using the lens formula $\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$:$\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{6.25} \implies \frac{1}{u_e} = -\frac{1}{25} - \frac{1}{6.25} = -\frac{1}{25} - \frac{4}{25} = -\frac{5}{25} = -\frac{1}{5}$.Thus,$|u_e| = 5 \, cm$.The distance between the two lenses is $L = |v_o| + |u_e| = 6 + 5 = 11 \, cm$.

In a compound microscope,the focal lengths of the two lenses are $1.5 \, cm$ and $6.25 \, cm$. If an object is placed at $2 \, cm$ from the objective lens and the final image is formed at $25 \, cm$ from the eye lens,the distance between the two lenses is ...... $cm$.

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