In a compound microscope,the focal lengths of | vedclass

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(A) For the objective lens,using the lens formula $\frac{1}{f_0} = \frac{1}{v_0} - \frac{1}{u_0}$:Given $f_0 = 2.5 \, cm$ and $u_0 = -3.75 \, cm$.$\fr

Vedclass · Accepted Answer

$11.67$

Vedclass · Answer

(A) For the objective lens,using the lens formula $\frac{1}{f_0} = \frac{1}{v_0} - \frac{1}{u_0}$:Given $f_0 = 2.5 \, cm$ and $u_0 = -3.75 \, cm$.$\frac{1}{2.5} = \frac{1}{v_0} - \frac{1}{-3.75} \Rightarrow \frac{1}{v_0} = \frac{1}{2.5} - \frac{1}{3.75} = \frac{1.5 - 1}{3.75} = \frac{0.5}{3.75} = \frac{1}{7.5}$.So,$v_0 = 7.5 \, cm$.For the eye lens,the image is formed at the least distance of distinct vision,$v_e = -25 \, cm$. Given $f_e = 5 \, cm$.Using $\frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e}$:$\frac{1}{5} = \frac{1}{-25} - \frac{1}{u_e} \Rightarrow \frac{1}{u_e} = -\frac{1}{25} - \frac{1}{5} = -\frac{6}{25}$.$u_e = -\frac{25}{6} \approx -4.167 \, cm$.The length of the microscope tube is $L = v_0 + |u_e| = 7.5 + 4.167 = 11.667 \, cm \approx 11.67 \, cm$.

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