When light travels from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle of incidence above which light is reflected only (no refraction occurs), and the value of this critical angle is given by
[tex]\theta_c = \arcsin ( \frac{n_2}{n_1} )[/tex]
where n2 is the refractive index of the second medium and n1 is the refractive index of the first medium.
In this problem, the first medium is the glass ([tex]n_1 = 1.50[/tex]), while the second medium is oil ([tex]n_2 =1.46[/tex]), therefore the critical angle is given by
[tex]\theta_c = \arcsin( \frac{1.46}{1.50} )=\arcsin(0.973)=76.7^{\circ}[/tex]
