Problem 17.39

พิจารณาแสงเดินทางจากน้ำไปยังอากาศ

จาก \(\frac{\sin \theta _{1}}{\sin \theta _{2}}=\frac{n_{2}}{n_{1}}\)

\(\frac{\sin \theta _{1}}{\sin (90º - 30º)}=\frac{1}{\frac{4}{3}}\)

\(\sin \theta _{1} =\frac{3\sin (60º)}{4}\)

\(\sqrt{1 - \cos ^{2}\theta _{1}}=\frac{3 3}{8}\)

\(\cos \theta _{1} =\frac{37}{8}\)

จาก \(\frac{n_{2}\cos \theta _{2}}{n_{1}\cos \theta _{1}}=\frac{s'}{s}\)

\(\frac{(1)\cos (90º - 30º)}{(\frac{4}{3})(\frac{37}{8})}=\frac{s'}{3.0 m}\)

\(s' = (3.0 m)(\frac{6\cos (60º)}{37})\)

\(s' = 1.48 m \;\blacksquare\)