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:
The topic of this question is Ray Optics and Optical Instruments.
We need to deduce the change in focal length after cutting a lens and apply that to the magnification formula of a compound microscope.
Step 2: Key Formula or Approach:
1. Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
2. Compound microscope magnification: $M \approx \frac{L}{f_o} \times \frac{D}{f_e}$.
Step 3: Detailed Explanation:
For a biconvex lens with equal radii $R$, the focal length $f$ is:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = \frac{2(\mu - 1)}{R} \]
When the lens is sliced vertically, we obtain a plano-convex lens ($R_1 = R, R_2 = \infty$). Its focal length $f'$ becomes:
\[ \frac{1}{f'} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{(\mu - 1)}{R} \]
Clearly, $f' = 2f$. The focal length of the objective lens has doubled.
Since the magnification formula contains the term $\frac{L}{f_o}$, doubling $f_o$ necessitates doubling the tube length $L$ to keep the magnification $M$ constant.
Step 4: Final Answer:
The tube length has to be increased two times.