A beam of light makes an angle of θi = 53.0˚ with the normal of a slab of transparent material of t=3.80 mm-thickness. The index of refraction of the glass is n2 = 1.39.
Angle 3 is 53.1
b) How far vertically (d) is the point at which the beam leaves the glass surface from the point at which the beam entered the glass? Provide the answer in mm.
To find the vertical distance (d) between the point at which the beam leaves the glass surface and the point at which it entered the glass, you can use the concept of refraction and Snell's Law.
Snell's Law states that the ratio of the sines of the angles of incidence (θi) and refraction (θr) is equal to the ratio of the indices of refraction (n1 and n2) of the two materials:
n1 * sin(θi) = n2 * sin(θr)
In this case, the incident angle (θi) is given as 53.0°, and the index of refraction of the glass (n2) is given as 1.39. We need to find the refracted angle (θr).
To find θr, we can rearrange Snell's Law:
sin(θr) = (n1 / n2) * sin(θi)
Plugging in the given values, we have:
sin(θr) = (1 / 1.39) * sin(53.0°)
Now, we can use the inverse sine function to find θr:
θr = sin^(-1)[(1 / 1.39) * sin(53.0°)]
Once we have θr, we can use basic trigonometry to find the vertical distance (d).
d = t * tan(θr)
Given that the thickness of the slab of transparent material (t) is 3.80 mm, we can substitute the values and calculate:
d = 3.80 mm * tan(θr)
Finally, using the calculated value for θr, we can find the vertical distance (d) in millimeters (mm).