JEE MAIN - Physics (2015 (Offline) - No. 6)
Assuming human pupil to have a radius of $$0.25$$ $$cm$$ and a comfortable viewing distance of $$25$$ $$cm$$, the minimum separation between two objects that human eye can resolve at $$500$$ $$nm$$ wavelength is :
$$100\,\mu m$$
$$300\,\mu m$$
$$1\,\mu m$$
$$30\,\mu m$$
Објашњење
$$\sin \theta = {{0.25} \over {25}} = {1 \over {100}}$$
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Resolving power $$ = {{1.22\lambda } \over {2\mu \sin \theta }} = 30\,\mu m.$$
Resolving power $$ = {{1.22\lambda } \over {2\mu \sin \theta }} = 30\,\mu m.$$