A thin light beam is shining onto a wall. You take a solid block of glass that is 3.0cm thick and place it between the wall and the beam. The refractive index of the glass is 1.5. The angle between the beam and the surface of the block is 60° What is the incident angle of the beam on the block.
the incident angle is measured relative to the normal to the surface
90º - 60º = 30º
To find the incident angle of the beam on the block, we can use the concept of Snell's Law. Snell's Law relates the incident angle (angle of the incoming beam with respect to the normal) and the refractive index of the two media involved.
The incident angle on the block can be found by reversing the roles of the glass and air. Let's consider the air as the initial medium and the glass as the final medium.
Snell's Law is given by:
n1 sinθ1 = n2 sinθ2
Where:
- n1 and n2 are the refractive indices of the initial and final media, respectively.
- θ1 is the incident angle of the light beam on the surface (with respect to the normal).
- θ2 is the angle of light inside the glass (with respect to the normal).
In this case, the incident angle on the block is given as 60°, and the refractive index of the glass is 1.5. We want to find the incident angle of the beam on the block.
Let's rearrange Snell's Law to solve for θ1:
θ1 = sin^(-1)((n2 * sinθ2) / n1)
Substituting the given values:
- n1 (refractive index of air) is approximately 1.
- n2 (refractive index of glass) is 1.5.
- θ2 (angle inside the glass) is 60° (since the incident angle and the angle inside the glass are the same).
θ1 = sin^(-1)((1.5 * sin(60°)) / 1)
Using a scientific calculator to compute this expression:
θ1 ≈ sin^(-1)(1.5 * 0.866) ≈ sin^(-1)(1.299) ≈ 51.91°
Therefore, the incident angle of the beam on the block is approximately 51.91°.