Ans with explanation 
 

Asked by Varsneya Srinivas | 20th Oct, 2017, 08:42: PM

Expert Answer:

 
From figure begin mathsize 12px style y over x space equals space tan theta space equals space fraction numerator sin begin display style theta end style over denominator square root of 1 minus sin squared theta end root end fraction space
f r o m space s n e l l s space l a w space mu subscript 0 sin 30 space equals space mu sin theta

sin c e space mu subscript 0 equals 1 comma space space sin theta space equals space fraction numerator 1 over denominator 2 mu end fraction

H e n c e space y over x space equals space fraction numerator begin display style bevelled fraction numerator 1 over denominator 2 mu end fraction end style over denominator square root of 1 minus begin display style bevelled fraction numerator 1 over denominator 4 mu squared end fraction end style end root end fraction space equals space fraction numerator 1 over denominator square root of 4 mu squared minus 1 end root end fraction equals space fraction numerator 1 over denominator square root of open parentheses 4 plus x close parentheses begin display style minus end style begin display style 1 end style end root end fraction space equals space fraction numerator 1 over denominator square root of open parentheses 3 plus x close parentheses end root end fraction

H e n c e space t r a j e c t o r y space i s space g i v e n space b y space y space equals space fraction numerator x over denominator square root of open parentheses 3 plus x close parentheses end root end fraction

a t space x space equals 1 comma space y space equals space fraction numerator 1 over denominator square root of 4 end fraction space equals space 1 half end style
Hence light ray is emerging at the point P which is 0.5 m above the reference line (dashed line in the figure)

Answered by  | 23rd Oct, 2017, 03:28: PM