We assume that the object is placed a little beyond its focal distance, say 2.5 cm, tan α = 2.54×10−1 = 254 Since , α is small. sin α ~ tan α = 254 ∴ Resolving limit of microscope, dx = 1μsinα1.22λ = 2×1×41.22×5500×10−10×25 = 2.0968 × 10−6cm ~ 2 × 10−6m
