Question

The numerical aperture of the objective of a microscope is 0.12. The limit of resolution, when the light of wavelength 6000$\mathop {\text{A}}\limits^0 $ is used to view an object is:
$
  {\text{A}}{\text{. 0}}{\text{.25}} \times {\text{1}}{{\text{0}}^{ - 7}}m \\
  {\text{B}}{\text{. 2}}{\text{.5}} \times {\text{1}}{{\text{0}}^{ - 7}}m \\
  {\text{C}}{\text{. 25}} \times {\text{1}}{{\text{0}}^{ - 7}}m \\
  {\text{D}}{\text{. 250}} \times {\text{1}}{{\text{0}}^{ - 7}}m \\
$

Answer 1

Hint: The limit of resolution of a microscope is equal to the ratio of the wavelength of incident light to two times the numerical aperture of the lens of the microscope. We have the values of wavelength and numerical aperture from which the limit of resolution can be found easily.

Formula used:
The limit of resolution of a microscope is given as
$\Delta d = \dfrac{\lambda }{{2NA}}$
where $\Delta d$ denotes the minimum distance that can be resolved, $\lambda $ is the wavelength of light used in the microscope, NA signifies the numerical aperture.

Detailed step by step solution:
The resolving power of a microscope can be defined as the ability of that microscope to distinguish between two very close objects. Mathematically, it is equal to the reciprocal of the separation between the two objects that the instrument is just able to resolve.

The limit of resolution of a microscope is defined as the minimum distance that the microscope can resolve. It is given as

$\Delta d = \dfrac{\lambda }{{2NA}}{\text{ }}...{\text{(i)}}$

The numerical aperture NA signifies how much light then the lens of the microscope can gather. Higher the value of numerical aperture better is the resolution of the microscope. Similarly, the smaller the wavelength better is the resolution.

We are given a microscope whose numerical aperture is

$NA = 0.12$

The wavelength of the incident light is given as

$\lambda = 6000\mathop {\text{A}}\limits^0 = 6 \times {10^{ - 7}}m$

Now by substituting these values in equation (i), we get

$
  \Delta d = \dfrac{{6 \times {{10}^{ - 7}}}}{{2 \times 0.12}} \\
   = 25 \times {10^{ - 7}}m \\
$

Hence, the correct answer is option C.

Note: The resolving power and limit of resolution are different quantities. The limit of resolution is the smallest measurable distance while the resolving power is defined as the reciprocal of this smallest measurable distance.