Question

A compound microscope has an objective of focal length $1.25\,\mathrm{cm}$ and an eyepiece of focal length $5\,\mathrm{cm}$ separated by a distance of $7.5\,\mathrm{cm}$. The total magnification produced by the microscope when the final image forms at infinity is

• A. 6.25
• B. 30
• C. 120
• D. 72.5

Hint:

Microscope Magnification (Infinity): $M = \frac{L}{f_o} \frac{D}{f_e}$. Microscope Magnification (Near Point): $M = \frac{L}{f_o} (1 + \frac{D}{f_e})$.

Answer 1

The Correct Option is B

Step 1: Understanding the Configuration:
Objective focal length $f_o = 1.25\,\mathrm{cm}$. Eyepiece focal length $f_e = 5\,\mathrm{cm}$. Separation distance (often approximated as tube length $L$ in competitive exams) $d = 7.5\,\mathrm{cm}$. Final image is at infinity (Normal adjustment).
Step 2: Formula for Magnification:
When the final image is at infinity, the magnification $M$ is: \[ M = \frac{L}{f_o} \times \frac{D}{f_e} \] where $D$ is the least distance of distinct vision ($25\,\mathrm{cm}$) and $L$ is the distance between the lenses (or tube length approximation used here).
Step 3: Calculation:
\[ M = \frac{7.5}{1.25} \times \frac{25}{5} \] \[ M = 6 \times 5 = 30 \] Note: While strictly $L$ in the formula is the distance between foci, in many exam contexts, the separation distance given is used directly as $L$ for the simplified formula.