The ratio of resolving powers of an optical microscope for two wavelengths $\lambda_1 = 4000 \mathring A$ and $\lambda_2 = 6000 \mathring A$ is :

Question

In a single slit diffraction pattern, a light of wavelength $ 6000 \, \text{Å} $ is used. The distance between the first and third minima in the diffraction pattern is found to be 3 mm when the screen is placed 50 cm away from the slits. The width of the slit is ______ $ \times 10^{-4} $ m.

Answer 1

The resolving power of an optical microscope is determined by its ability to distinguish between two closely spaced objects. The resolving power $ R $ is inversely proportional to the wavelength $\lambda$ used for observation. Mathematically, it can be given by:

R \propto \frac{1}{\lambda}

Given two wavelengths, $\lambda_1 = 4000 \mathring{A}$ and $\lambda_2 = 6000 \mathring{A}$, the ratio of resolving powers $R_1$ to $R_2$ is:

\frac{R_1}{R_2} = \frac{\lambda_2}{\lambda_1}

Substituting the given values:

\frac{R_1}{R_2} = \frac{6000}{4000} = \frac{3}{2}

Thus, the ratio of the resolving powers is 3:2, which corresponds to the correct answer from the given options.

Conclusion: The correct answer is 3:2. This choice is justified because resolving power increases with a decrease in the wavelength, thus for smaller wavelength ($4000 \mathring{A}$), the resolving power is higher.

Answer 2

The resolving power of an optical microscope is determined by its ability to distinguish between two closely spaced objects. The resolving power $ R $ is inversely proportional to the wavelength $\lambda$ used for observation. Mathematically, it can be given by:

R \propto \frac{1}{\lambda}

Given two wavelengths, $\lambda_1 = 4000 \mathring{A}$ and $\lambda_2 = 6000 \mathring{A}$, the ratio of resolving powers $R_1$ to $R_2$ is:

\frac{R_1}{R_2} = \frac{\lambda_2}{\lambda_1}

Substituting the given values:

\frac{R_1}{R_2} = \frac{6000}{4000} = \frac{3}{2}

Thus, the ratio of the resolving powers is 3:2, which corresponds to the correct answer from the given options.

Conclusion: The correct answer is 3:2. This choice is justified because resolving power increases with a decrease in the wavelength, thus for smaller wavelength ($4000 \mathring{A}$), the resolving power is higher.

The ratio of resolving powers of an optical microscope for two wavelengths $\lambda_1 = 4000 \mathring A$ and $\lambda_2 = 6000 \mathring A$ is :

The Correct Option is B

Solution and Explanation

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