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JEE Mains · Physics · STD 12 - 10. Wave optics
The diameter of the objective lens of microscope makes an angle \(\beta \) at the focus of the microscope. Further, the medium between the object and the lens is an oil of refractive index \(n\). Then the resolving power of the microscope
- A increases with decreasing value of \(n\)
- B increases with decreasing value of \(\beta \)
- C increases with increasing value of \(n\, sin\, 2\beta \)
- D increases with increasing value of \(\frac{1}{{n\,\sin \,2\beta }}\)
Answer & Solution
Correct Answer
(C) increases with increasing value of \(n\, sin\, 2\beta \)
Step-by-step Solution
Detailed explanation
Resolving power of microscope, \(\mathrm{R.P.}=\frac{2 \mathrm{n} \sin \theta}{\lambda}\) \(\lambda=\) Wavelength of light used to illuminate the object \(\mathrm{n}=\) Refractiveindex of the medium between object and objective \(\theta=\) Angle
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