Tewtor.ai

You’re very close. Let’s tidy the idea and then fill the table.

Key idea (simple rule):

• Total magnification $M = (\text{ocular}) \times (\text{objective})$.
• Field of view (FOV) scales inversely with magnification: $FOV \times M = \text{constant}$.
• So if magnification goes up by a factor $k$, FOV goes down by the same factor $k$.

Your numbers:

• With $10\times$ ocular:
  • Objective $4\times$ → total $40\times$.
  • Objective $10\times$ → total $100\times$.
  • Objective $40\times$ → total $400\times$.

Relationship column (how many times bigger the higher magnification is compared to the first row at $40\times$):

• Row 1 ($40\times$ vs $40\times$): factor $1$.
• Row 2 ($100\times$ vs $40\times$): factor $100/40 = 2.5$.
• Row 3 ($400\times$ vs $40\times$): factor $400/40 = 10$.

Now FOV values: We need one known FOV to anchor the rest. Your manual seems to give $FOV$ at $40\times$ as $4\,\text{mm}$ (that’s what you used). If that’s the given, then:

• Row 1 ($40\times$): $FOV = 4\,\text{mm} = 4000\,\mu\text{m}$.
• Row 2 ($100\times$): divide by $2.5$ → $FOV = \frac{4}{2.5} = 1.6\,\text{mm} = 1600\,\mu\text{m}$.
• Row 3 ($400\times$): divide by $10$ → $FOV = \frac{4}{10} = 0.4\,\text{mm} = 400\,\mu\text{m}$.

So yes, your calculation for the second row ($1.6\,\text{mm}$ and $1600\,\mu\text{m}$) is right, given $4\,\text{mm}$ at $40\times$.

Quick check question: Does your lab manual explicitly give the $FOV$ at $40\times$ as $4\,\text{mm}$? If it gives a different starting FOV, we’ll scale all rows from that number using the same ratios.

What was the key insight here? That $FOV \times \text{magnification}$ stays constant, so FOV changes by the inverse of the magnification factor.