A compound microscope consists of an objectiv | vedclass

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Question

(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 $n = 50$.

Similar Questions

In a compound microscope,the objective lens and eyepiece have focal lengths of $0.95 \ cm$ and $5 \ cm$ respectively,and are kept at a distance of $20 \ cm$. The final image is formed at a distance of $25 \ cm$ from the eyepiece. Calculate the magnifying power.

If the focal length of the objective and eye lens are $1.2 \, cm$ and $3 \, cm$ respectively,and the object is placed $1.25 \, cm$ away from the objective lens,and the final image is formed at infinity,the magnifying power of the microscope is:

The magnifying power of a simple microscope is $6$. The focal length of its lens in metres will be,if the least distance of distinct vision is $25\,cm$.

If the focal length of the objective lens and the eye lens are $4 \, mm$ and $25 \, mm$ respectively in a compound microscope,and the length of the tube is $16 \, cm$,find its magnifying power for the relaxed eye position.

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