Light enters a plate of borosilicate crown glass
at an angle of 40.2 degrees � with respect to the nor-
mal to the air-glass surface. If the index of
refraction of the glass is 1.52, find the angle
made by the refracted beam with the normal.
Answer in units of �.
Snell's law says that
sin 40.2 = 1.52 sin(theta)
theta is the angle of refraction.
Solve for it.
sin(theta) = 0.4246
theta = 25.1 degrees
To find the angle made by the refracted beam with the normal, we can use Snell's law, which relates the angles of incidence and refraction to the indices of refraction of the two media involved.
Snell's law states: n1 * sin(θ1) = n2 * sin(θ2)
Where:
- n1 and n2 are the indices of refraction of the two media
- θ1 is the angle of incidence
- θ2 is the angle of refraction
In this case, the light is entering the borosilicate crown glass from air. The index of refraction of air is approximately 1 (close to vacuum), and the index of refraction of the glass is given as 1.52.
Given:
- n1 = 1 (for air)
- n2 = 1.52 (for borosilicate crown glass)
- θ1 = 40.2 degrees
Let's plug these values into Snell's law and solve for θ2:
1 * sin(40.2 degrees) = 1.52 * sin(θ2)
To find the value of θ2, we rearrange the equation:
sin(θ2) = (1 * sin(40.2 degrees)) / 1.52
θ2 = arcsin((1 * sin(40.2 degrees)) / 1.52)
Now, we can calculate the value of θ2 using a calculator with trigonometric functions:
θ2 ≈ arcsin((1 * sin(40.2 degrees)) / 1.52)
The result will be the angle made by the refracted light beam with the normal to the air-glass surface, expressed in degrees.