Respuesta :
Answer:
The ratio of the refractive indices of the liquids is 1.22
Solution:
Critical angle for A and B interface, [tex]\theta_{AB} = 38.0^{\circ}[/tex]
Critical angle for A and B interface, [tex]\theta_{AC} = 49.5^{\circ}[/tex]
Now,
From the relation in between the critical angle and refractive index:
[tex]n = \frac{1}{\theta_{c}}[/tex] (1)
where
n = rafractive index
[tex]\theta_{c}[/tex] = critical angle
Thus
For AB interface:
[tex]n_{B} = \frac{1}{sin\theta_{AB}} = \frac{1}{sin(38.0^{\circ})} = 1.606[/tex]
For AC interface:
[tex]n_{C} = \frac{1}{sin\theta_{AC}} = \frac{1}{sin(49.5^{\circ})} = 1.315[/tex]
Thus the ratio of the refractive indices of these liquids can be given as:
[tex]\frac{n_{B}}{n_{C}} = \frac{1.606}{1.315}[/tex]
[tex]\frac{n_{B}}{n_{C}} = 1.22[/tex]
