Question

A ray in glass is incident onto a water-glass interface, at an angle of incidence equal to half the critical angle for that interface. The indices of refraction for water and the glass are 1.33 and 1.88, respectively. The angle that the refracted ray in the water makes with the normal is closest to:

a) 23 degrees
b) 33 degrees
c) 28 degrees
d) 38 degrees
e) 43 degrees

Answer 1

to find the citical angle first

$A_c=sin^{-1}(\frac{n_{water}}{n_{glass}})=sin^{-1}(\frac{1.33}{1.88})=45.02^{o}$

now

    $A_{glass}=\frac{A_c}{2}=\frac{45.02}{2}=22.51^{o}$

we can say that

$n_{glass}*sin(A_{glass})=n_{water}*sin(A_{water})$

$1.88*sin(22.51)=1.33*sin(A_{water})$

we have to clear A water the

$A_{water}=32.7629^{0}$

$A_{water}\approx 33$