A light ray is directed at a vertical face of a glass cube. The angle of incidence at the vertical face is 39° and the angle of refraction is 25° as shown in the diagram.Show that the refractive index of the glass is about 1.5.
Always check that the angle of incidence and refraction are the angles between the normal and the light ray. If the angle between the light ray and the boundary is calculated instead, calculate 90 – θ (since the normal is perpendicular to the boundary) to get the correct angle
The angle of incidence is greater than the critical angle and the incident refractive index n1 is greater than the refractive index of the material at the boundary n2
A glass cube is held in contact with a liquid and a light ray is directed at a vertical face of the cube.The angle of incidence at the vertical face is 39° and the angle of refraction is 25° as shown in the diagram.The light ray is totally internally reflected at X.(i) Complete the diagram to show the path of the ray beyond X to the air and calculate the critical angle for the glass-liquid boundary.(ii) Calculate the refractive index of the liquid.
(i)Step 1: Draw the reflected angle at the glass-liquid boundary
Step 2: Draw the refracted angle at the glass-air boundary
Step 3: Calculate the critical angle
(ii)Step 1: Write down the known quantities
Step 2: Write out Snell’s Law or the equation for critical angle
Step 3: Calculate the refractive index of the liquid