A compound microscope consists of an objectiv | vedclass

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(A) For minimum strain on the eye,the final image is formed at infinity.For the eyepiece,the object must be placed at its focal length,so $u_e = -f_e

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(A) For minimum strain on the eye,the final image is formed at infinity.For the eyepiece,the object must be placed at its focal length,so $u_e = -f_e = -5 \, cm$.The separation between the lenses is $L = 10 \, cm$. The image formed by the objective lens $(v_o)$ acts as the object for the eyepiece.Thus,$v_o = L - |u_e| = 10 \, cm - 5 \, cm = 5 \, cm$.Using the lens formula for the objective lens: $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$.Substituting the values: $\frac{1}{5} - \frac{1}{u_o} = \frac{1}{1} \Rightarrow -\frac{1}{u_o} = 1 - \frac{1}{5} = \frac{4}{5}$.Therefore,$u_o = -\frac{5}{4} \, cm$. The distance is $|u_o| = \frac{5}{4} \, cm = \frac{50}{40} \, cm$.Comparing this with $\frac{n}{40} \, cm$,we get $n = 50$.

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