Answer:
The angle of refraction is closest to
![45^(o)](https://img.qammunity.org/2021/formulas/physics/college/9bp4vd5siz18h6fbp6o2kcdshtvisgxc9u.png)
Step-by-step explanation:
Snell's law compares the ratios of the angles of incident and refraction, and it would be applied in solving this problem.
Given the
Refractive index η= 1.42
angle of incident = i
angle of refraction = r = 1/2 x i = i/2
applying Snell's law;
η =
![(sini)/(sin(i)/(2) )](https://img.qammunity.org/2021/formulas/physics/college/q7pau3wvvk7orko8ecq3q7m33j528lz9vk.png)
applying trigonometric identity (sin 2x=2sinxcosx )
sin 2i = 2sinicosi
1.42 =
![(2 sin(i/2)cos(i/2))/(sin (i/2) )](https://img.qammunity.org/2021/formulas/physics/college/djeaeyuaoy33uda4ql35vq3wfqpt3tojn0.png)
cos i/2 = 1.42/2
cos i/2 = 0.71
i/2 =
0.71 =
![44.765^(o)](https://img.qammunity.org/2021/formulas/physics/college/vxg9x8favcqtswy167kpygqk2h40n7j5on.png)
i/2 ≈
![45^(o)](https://img.qammunity.org/2021/formulas/physics/college/9bp4vd5siz18h6fbp6o2kcdshtvisgxc9u.png)
Therefore the angle of refraction is closest to
![45^(o)](https://img.qammunity.org/2021/formulas/physics/college/9bp4vd5siz18h6fbp6o2kcdshtvisgxc9u.png)