Question

A compound microscope is designed with two symmetric biconvex lenses. The objective lens is cut vertically, creating two identical plano-convex lenses. One of them is used in place of original objective lens. To retain same magnification keeping the object distance unchanged, the tube length has to be:

• A. increased two times
• B. increased 3/2 times
• C. decreased two times
• D. decreased 3/2 times

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to determine how the focal length of the objective lens changes when it is cut in half and how this affects the required tube length to maintain constant magnification.
Step 2: Detailed Explanation:
For a symmetric biconvex lens with radius $R$, the focal length $f$ is given by:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = \frac{2(\mu - 1)}{R} \]
When the lens is cut vertically into two plano-convex lenses, for each plano-convex lens ($R_1 = R, R_2 = \infty$):
\[ \frac{1}{f'} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{(\mu - 1)}{R} \]
Comparing the two, we see $f' = 2f$. The focal length of the objective lens doubles.
The magnification of a compound microscope is approximately given by:
\[ M \approx \frac{L}{f_o} \times \frac{D}{f_e} \]
where $L$ is the tube length.
To keep $M$ the same while $f_o$ becomes $2f_o$, the tube length $L$ must also be doubled ($L' = 2L$) to maintain the ratio.
Step 3: Final Answer:
The tube length has to be increased two times.