Answer:
C ) 1.53
Explanation:
The critical angle of a material is given by the formula
[tex]sin c = \frac{1}{n}[/tex]
where
c is the critical angle
n is the refractive index
This formula is valid if the second medium is air (which is the case of the problem).
In this problem, we know the critical angle:
[tex]c=40.8^{\circ}[/tex]
Therefore we can rearrange the equation to find the refractive index:
[tex]n=\frac{1}{sin c}=\frac{1}{sin 40.8^{\circ}}=1.53[/tex]
