If a microscope is placed in air, the minimum separation of two objects seen as distinct is 6 pm . If the…
If a microscope is placed in air, the minimum separation of two objects seen as distinct is 6 pm . If the same is placed in a medium of refractive index 1.5 , then the minimum separation of the two objects to see as distinct is
$4 \mu \mathrm{~m}$
$6 \mu \mathrm{~m}$
$3 \mu \mathrm{~m}$
$9 \mu \mathrm{~m}$
Solution
In air, $\mathrm{d}=6 \mu \mathrm{~m}, \mathrm{n}=1.5$ In medium, $\mathrm{d}^{\prime}=\frac{\mathrm{d}}{\mathrm{n}}=\frac{6}{1.5}=4 \mu \mathrm{~m}$