Question

In an optical instrument- microscope, the wavelength of light used are \(λ_1 = 6800\) Å and \(λ_2 = 5100\) Å. The ratio of the resolving power of microscope corresponding to '\(λ_1\)' to that for '\(λ_2\)' is

• A. \(9:16\)
• B. \(5:4\)
• C. \(4:3\)
• D. \(3:4\)

Hint:

Resolving power of a microscope is proportional to 1 / lambda.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The resolving power of a microscope is its ability to show two close points as separate. A shorter wavelength gives better resolving power.

Step 2: Key Formula or Approach:
\[ \text{R.P.} = \frac{2\mu\sin\theta}{1.22\lambda} \Rightarrow \text{R.P.}\propto\frac1\lambda \]

Step 3: Detailed Explanation:
\[ \frac{(\text{R.P.})_1}{(\text{R.P.})_2} = \frac{\lambda_2}{\lambda_1} = \frac{5100}{6800} = \frac34 \]
So the resolving power at 6800 Å is three fourths of the resolving power at 5100 Å. Option (C) 4:3 is the inverse ratio, which would be right if the resolving power was proportional to \(\lambda\). Options (A) 9:16 and (B) 5:4 do not match.

Final Answer:
The ratio of resolving powers is 3:4, option (D). \[ \boxed{3:4 \text{ (D)}} \]