Question

Two points separated by a distance of \[0.1mm\] can just be inspected in a microscope when the light of a wavelength \[600{A^ \circ }\] is used. If the light of wavelength \[4800{A^ \circ }\] is used, the limit of resolution will become
a. 0.80 mm
b. 0.12 mm
c. 0.10 mm
d. 0.08 mm

Answer 1

Hint: The resolution power of a microscope is defined as its ability to distinguish between any two lines placed close to each other in an object. It is inversely proportional to the distance between the two objects to be distinguished. It depends on the wavelength and aperture of the lens. Using these factors, find the limit of resolution.

Complete step by step answer:
Step I:
The resolution power of a lens depends on its wavelength of incident light. If the wavelength is decreased, then the power of resolution increases. On the other hand, if the wavelength of incident light is increased then the power of resolution will decrease.
Also it varies with the aperture of the lens. If the aperture of the lens is increased, then the resolving power of the lens also increases. But if the aperture of the lens is decreased, then the resolving power also decreases.
 Let \[\lambda \] be the wavelength and let ‘a’ be the aperture of the lens used in the microscope.
Let ‘x’ be the resolving power. Use the relation \[x \propto \dfrac{{2a}}{\lambda }\]
From above relation it is clear that \[x \propto a\] and \[x \propto \dfrac{1}{\lambda }\]

Step II:
Let \[{\lambda _1} = 600{A^ \circ }\],then \[{x_1} = 0.1m\]
and \[{\lambda _2} = 4800{A^ \circ }\]then \[{x_2} = ?\]

Step III:
Taking ratio of both wavelengths and powers, \[\dfrac{{{x_1}}}{{{x_2}}} = \dfrac{{{\lambda _1}}}{{{\lambda _2}}}\]
${x_2} = \dfrac{{{x_1}{\lambda _2}}}{{{\lambda _1}}}$
Substituting all the values and evaluating the value of resolution,
\[{x_2} = \dfrac{{4800 \times 0.1}}{{6000}}\]
\[{x_2} = \dfrac{{48}}{{600}}\]
\[{x_2} = 0.08mm\]
The limit of resolution will become $0.08mm$.

Hence, the correct answer is option (D).

Note: Resolving power is also affected by the refractive index of the medium. It varies directly with the refractive index. If the refractive index of the medium increases, the power of resolution also increases. But if the index decreases, then power of resolution also decreases.