In an optical instrument- microscope, the wavelengths of light used are $\lambda_1 = 6800 \text{ Å}$ and $\lambda_2 = 5100 \text{ Å}$. The ratio of the resolving power of the microscope corresponding to $\lambda_1$ to that for $\lambda_2$ is
- A$9 : 16$
- B$5 : 4$
- C$4 : 3$
- D$3 : 4$
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Similar Questions
The numerical aperture of a microscope is $0.12$, and the wavelength of light used is $600 \, nm$. Then its limit of resolution will be nearly $............. \, \mu m$.
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View SolutionTwo point white dots are $1 \ mm$ apart on a black paper. They are viewed by an eye with a pupil diameter of $3 \ mm$. Approximately,what is the maximum distance at which the dots can be resolved by the eye? (Take wavelength of light $= 500 \ nm$)
MediumAIIMS 2008
View SolutionWavelengths of light used in an optical instrument are $\lambda_1 = 4000 \; \mathring{A}$ and $\lambda_2 = 5000 \; \mathring{A}$. The ratio of their respective resolving powers (corresponding to $\lambda_1$ and $\lambda_2$) is:
Large apertures of telescopes are used for
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