Tube Length Adjustment for Compound Microscope Magnification

To determine how the tube length of the compound microscope must be adjusted, we analyze the change in the focal length of the objective lens and its effect on the magnification of the microscope.

1. Change in Focal Length of the Objective Lens

For a symmetric biconvex lens of refractive index $\mu$ with radii of curvature $R_1 = R$ and $R_2 = -R$, the Lens Maker's Formula gives its initial focal length $f_o$: $\frac{1}{f_o} = (\mu - 1) \left( \frac{1}{R} - \left(-\frac{1}{R}\right) \right) = \frac{2(\mu - 1)}{R}$

When the lens is cut vertically along the plane perpendicular to the optical axis, it is split into two identical plano-convex lenses. For one such plano-convex lens, the radii are $R_1 = R$ and $R_2 = \infty$. Its new focal length $f_o'$ is given by: $\frac{1}{f_o'} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{\mu - 1}{R}$

Comparing the two focal lengths: $f_o' = 2 f_o$

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2. Tube Length Adjustment for Same Magnification

The total magnification $M$ of a compound microscope is given by: $M = m_o \times m_e \approx -\frac{L}{f_o} \times m_e$ where:

• $L$ is the tube length of the microscope,
• $f_o$ is the focal length of the objective lens,
• $m_e$ is the magnification of the eyepiece.

To retain the same total magnification $M$ while keeping the eyepiece parameters and object distance fixed, the ratio $\frac{L}{f_o}$ must remain constant: $\frac{L'}{f_o'} = \frac{L}{f_o}$

Substituting $f_o' = 2f_o$: $\frac{L'}{2f_o} = \frac{L}{f_o} \implies L' = 2L$

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Conclusion

The new tube length $L'$ is twice the original tube length $L$. Thus, the tube length has to be increased two times.

Correct Option: A