JEE Main 2014 · 19 Apr · Shift 1 · Physics Q24

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The diameter of the objective lens of microscope makes an angle β 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.

Increases with decreasing value of n

Increases with decreasing value of β

Increases with increasing value of

n\sin2\beta

Increases with increasing value of

\dfrac1{n\sin2\beta}

Answer and worked solution

Official answer: C

The resolving power of a microscope is $\dfrac{2n\sin\theta}{1.22\lambda}$ , where θ is the half-angle subtended by the objective and $n\sin\theta$ is the numerical aperture. It grows with $n$ and with the angle; of the options, it increases with increasing $n\sin2\beta$ , as keyed.

Topics:Resolving power Optical instruments

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JEE Main 2014 · 19 Apr · Shift 1, Physics question 24