Given:
The refractive index of water is n1 = 1.33
The critical angle is
![\theta=69.3\text{ degrees}](https://img.qammunity.org/2023/formulas/physics/college/6q2grbjmu3rzbqh0pgluuvtfhmol55zrna.png)
To find the refractive index of the substance.
Step-by-step explanation:
The refractive index of the substance can be calculated as
![\begin{gathered} n2\text{ =}\frac{n1}{sin\text{ }\theta} \\ =(1.33)/(sin(69.3^(\circ))) \\ =\text{ 1.42} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/qrqmla8dxjt0cwzu9km9bfxz1rx9d75j35.png)
Thus, the refractive index of the substance is 1.42