A beam of light makes an angle of 13.0° with the normal of a mirror made of 4.70-mm-thick glass silvered on the back. If the index of refraction of the glass is 1.30, how far is the point at which the beam leaves the glass surface (after being reflected from the silver backing) from the point at which the beam entered the glass? Provide the answer in mm.
To find the distance at which the beam leaves the glass surface, we can use the concept of refraction and Snell's law.
First, we need to determine the angle of incidence. The angle of incidence is the angle between the incident beam and the normal to the surface. In this case, the angle of incidence is equal to 13.0°.
Next, we can use Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the indices of refraction of the two media.
sin(angle of incidence) / sin(angle of refraction) = n2 / n1
In this case, the incident medium is air and the refractive index of air is approximately 1.00. The refractive index of the glass is given as 1.30.
sin(13.0°) / sin(angle of refraction) = 1.30 / 1.00
Now, rearranging the equation to solve for the angle of refraction:
sin(angle of refraction) = (sin(13.0°) / 1.30) [dividing both sides by 1.30]
Using the inverse sine function (sin^-1), we can find the angle of refraction:
angle of refraction = sin^-1(sin(13.0°) / 1.30)
After finding the angle of refraction, we can calculate the distance:
Distance = thickness of the glass / tan(angle of refraction)
Given that the thickness of the glass is 4.70 mm, we can substitute this value into the equation:
Distance = 4.70 mm / tan(angle of refraction)
Calculating the final value will give you the distance at which the beam leaves the glass surface.