What will be the limit of resolution of a microscope if its numerical aperture is 0.12 and the wavelength of the light used is 6000 $A^0$ (numerical aperture = $\mu \sin 0$ )?

The correct option is D $30.5$ $\times$ ${10}^{-7}$ m

Given $\lambda =6000A^{\circ }=6000\times {10}^{-10}m=6\times {10}^{-7}m$

Numerical aperture, $NA=0.12$

Limit of resolution is given as

$X=\frac{1.22\lambda }{2NA}$

$\therefore X=\frac{1.22\times 6\times {10}^{-7}}{2\times 0.12}=30.5\times {10}^{-7}m$

Hence the correct answer is option (B).