Question:
A compound microscope consists of an objective lens of focal length 1 cm and an eyepiece of focal length 5 cm with a separation of 10 cm.
The distance between an object and the objective lens, at which the strain on the eye is minimum is \(\dfrac{n}{40}\) cm. The value of \(n\) is _____.
A compound microscope consists of an objective lens of focal length 1 cm and an eyepiece of focal length 5 cm with a separation of 10 cm.
The distance between an object and the objective lens, at which the strain on the eye is minimum is \(\dfrac{n}{40}\) cm. The value of \(n\) is _____.
The distance between an object and the objective lens, at which the strain on the eye is minimum is \(\dfrac{n}{40}\) cm. The value of \(n\) is _____.
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Minimum strain means the final image is at infinity, so the intermediate image lies at the eyepiece focus. Then use the lens formula for the objective.
Updated On: Nov 5, 2026
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Correct Answer: 50
Solution and Explanation
Strain on the eye is minimum when the final image is formed at infinity, so the intermediate image lies at the focus of the eyepiece: distance from eyepiece = 5 cm.
Separation of lenses = 10 cm, so the image formed by the objective is at \(v = 10 - 5 = 5\) cm from the objective.
Lens formula for the objective (\(f = 1\) cm): \(\dfrac1v - \dfrac1u = \dfrac1f\).
\(\dfrac15 - \dfrac1u = 1 \Rightarrow \dfrac1u = \dfrac15 - 1 = -\dfrac45\).
\(u = -\dfrac54\) cm, so the object distance is \(1.25\) cm \(= \dfrac{50}{40}\) cm.
\(n = 50\).
Answer: 50
Separation of lenses = 10 cm, so the image formed by the objective is at \(v = 10 - 5 = 5\) cm from the objective.
Lens formula for the objective (\(f = 1\) cm): \(\dfrac1v - \dfrac1u = \dfrac1f\).
\(\dfrac15 - \dfrac1u = 1 \Rightarrow \dfrac1u = \dfrac15 - 1 = -\dfrac45\).
\(u = -\dfrac54\) cm, so the object distance is \(1.25\) cm \(= \dfrac{50}{40}\) cm.
\(n = 50\).
Answer: 50
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