JEE Mains 2020 Model Paper 1 New Pattern

We assume that the object is placed a little beyond its focal distance, say 2.5 cm,
tan α =

4×10−1

2.5

=

4

25

Since , α is small.
sin α ~ tan α =

4

25

∴ Resolving limit of microscope,
dx =

1.22λ

1µsinα

=

1.22×5500×10−10×25

2×1×4

= 2.0968 × 10−6cm
~ 2 × 10−6m