Question

A circular beam of light of diameter 2cm fall on a plane surface of glass.

Answer 1

Let d be diameter of refracted beam. Then,

d=PQcos60∘d=PQcos⁡60∘

and d′=PQcosrd′=PQcos⁡r

ie d′d=cosrcos60∘d′d=cos⁡rcos⁡60∘

=2cosr=2cos⁡r

or d′=2dcosrd′=2dcos⁡r


sinr=siniμsin⁡r=sin⁡iμ

=3√232=3232

=13–√=13

∴cosr=1−sin2r−−−−−−−−√=23−−√∴cos⁡r=1−sin2⁡r=23

∴d′=(2)(2)23−−√=423−−√∴d′=(2)(2)23=423cmcm

=3.26cm=3.26cm

Hence d is the correct answer.

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