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(A) The resolving power of a microscope is given by $R = \frac{1}{d} = \frac{2n \sin \beta}{1.22 \lambda}$,where $d$ is the minimum separation between

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$4$

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(A) The resolving power of a microscope is given by $R = \frac{1}{d} = \frac{2n \sin \beta}{1.22 \lambda}$,where $d$ is the minimum separation between two objects.Thus,the minimum separation $d$ is inversely proportional to the refractive index $n$ of the medium $(d \propto \frac{1}{n})$.Given for air $(n_1 = 1)$,$d_1 = 6 \mu m$.For a medium with refractive index $n_2 = 1.5$,the new minimum separation $d_2$ is calculated as:$d_2 = \frac{d_1 	imes n_1}{n_2} = \frac{6 \mu m 	imes 1}{1.5} = 4 \mu m$.Therefore,the minimum separation is $4 \mu m$.

If a microscope is placed in air,the minimum separation of two objects seen as distinct is $6 \mu m$. If the same is placed in a medium of refractive index $1.5$,then the minimum separation of the two objects to be seen as distinct is: (in $\mu m$)

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