Answer:
[tex]\theta_{r}[/tex]= 48.98°≅49°
Explanation:
refractive index of liquid n_{liquid}= 1.398
refractive index of glass n_{glass}= 1.541
angle of incidence i= 43.2°
we have to find corresponding angle of refraction
by snell's law we write
[tex]n_{glass}sin\theta_i=n_{water}sin\theta_r[/tex]
now putting values
[tex]1.541sin43.2=1.398sin\theta_r[/tex]
θ_{r}= [tex]sin^{-1}(\frac{1.541sin(43.2)}{1.398} )[/tex]
on calculating we get
[tex]\theta_{r}[/tex]= 48.98°≅49°
