A beam of light makes an angle of θi = 7.0˚ with the normal of a slab of transparent material of t=1.00-mm-thickness. If the index of refraction of the glass is n2 = 1.40 , 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) traveled by the beam inside the glass slab, you can use the laws of refraction.
The law of refraction, also known as Snell's law, states:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
Where:
- n₁ is the index of refraction of the initial medium (in this case, air),
- n₂ is the index of refraction of the final medium (in this case, glass),
- θ₁ is the angle of incidence (the angle between the incident ray and the normal),
- θ₂ is the angle of refraction (the angle between the refracted ray and the normal).
In this problem, the angle of incidence (θ₁) is given as 7.0˚, the index of refraction of the glass (n₂) is given as 1.40, and we need to find the vertical distance (d) traveled by the beam inside the glass slab.
To calculate θ₂, we need to use Snell's law:
sin(θ₂) = (n₁ / n₂) * sin(θ₁)
Given that n₁ (the index of refraction of air) is approximately 1.00 and θ₁ (the angle of incidence) is 7.0˚, we can calculate sin(θ₂):
sin(θ₂) = (1.00 / 1.40) * sin(7.0˚)
Now, find the value of sin(θ₂).