\(tan \theta_c=\frac{\frac{7.6cm}{4}}{2.25cm}=\frac{1.9}{2.25}\)

\(\theta_c=tan^{-1}\frac{1.9}{2.25}=40.18^0\)

\(n=\frac{1}{sin \theta_c}=\frac{1}{sin40.18^0}=1.55\)