AIIMS 2019 Physics Microscope MCQ Question

Option B correctly calculates the resolving power (RP) of a microscope using the formula $RP = \frac{2\mu \sin \theta}{1.22 \lambda}$, where $\mu$ is the refractive index (assumed to be 1 for air), $\theta$ is the angle of the lens, and $\lambda$ is the wavelength of light (given as $6 \times 10^{-7}$ m). Substituting $\sin \theta$ with $0.2$ yields an RP of approximately $10.9 \times 10^5 \, \text{m}^{-1}$, demonstrating the microscope's ability to distinguish between closely spaced objects.

Option A incorrectly equates $\tan \theta$ and $\sin \theta$ without considering the small angle approximation, which is only valid for small angles. Options C and D are not applicable, as they do not provide relevant information or calculations. Understanding the relationship between angle, wavelength, and resolving power is crucial for optimizing microscope performance.