Refractive index

The light passes through the interface between air and glass with a refractive index of 1.5. Find:
(a) the angle of refraction if light strikes the interface from the air at an angle of 40°.
(b) the angle of refraction when light hits the glass interface at an angle of 40°.
(c) the angle of incidence if the light is refracted at 20° on impact from the glass.
(d) cut-off angle (justify for which way the light passes through the interface)

Final Answer:

A =  25.374 °
B =  74.6186 °
C =  30.8659 °
D =  41.8103 °

Step-by-step explanation:

$$\begin{gathered} n=1.5 \\ \alpha =40{}^{\circ } \\ \sin \alpha :\sin A=n \\ {A}_1=\arcsin (\sin \alpha /n)=\arcsin (\sin 40°/n)=\arcsin (\sin 40°/1.5)=\arcsin (0.642788/1.5)=0.44286 \\ A=A_1\to \text{°}=A_1\cdot \frac{180}{\pi }\text{°}=0.4429\cdot \frac{180}{\pi }\text{°}=25.374\text{°}=25.374^{\circ }=25°22^{\prime }26" \end{gathered}$$

$$\begin{gathered} \beta =40{}^{\circ } \\ \sin B:\sin \beta =n \\ {B}_1=\arcsin (n\cdot \sin \beta )=\arcsin (n\cdot \sin 40°)=\arcsin (1.5\cdot \sin 40°)=\arcsin (1.5\cdot 0.642788)=1.30234 \\ B=B_1\to \text{°}=B_1\cdot \frac{180}{\pi }\text{°}=1.3023\cdot \frac{180}{\pi }\text{°}=74.619\text{°}=74.6186^{\circ }=74°37^{\prime }7" \end{gathered}$$

$$\begin{gathered} \gamma =20{}^{\circ } \\ \sin C:\sin \gamma =n \\ {C}_1=\arcsin (n\cdot \sin \gamma )=\arcsin (n\cdot \sin 20°)=\arcsin (1.5\cdot \sin 20°)=\arcsin (1.5\cdot 0.34202)=0.53871 \\ C=C_1\to \text{°}=C_1\cdot \frac{180}{\pi }\text{°}=0.5387\cdot \frac{180}{\pi }\text{°}=30.866\text{°}=30.8659^{\circ }=30°51^{\prime }57" \end{gathered}$$

$$\begin{gathered} \delta =90{}^{\circ } \\ \sin \delta :\sin D=n \\ {D}_1=\arcsin (\sin \delta /n)=\arcsin (\sin 90°/n)=\arcsin (\sin 90°/1.5)=\arcsin (1/1.5)=0.72973 \\ D=D_1\to \text{°}=D_1\cdot \frac{180}{\pi }\text{°}=0.7297\cdot \frac{180}{\pi }\text{°}=41.81\text{°}=41.8103^{\circ }=41°48^{\prime }37" \end{gathered}$$

Try another example