A point source of light is 96.7 cm below the surface of a body of water. Find the diameter of the circle at the surface through which light emerges from the water. Water has an index of refraction of 1.33.
There is a maximum angle that a ray from the source can have with the vertical, and not be totally reflected. That angle is called the critical angle, theta. It is giben by the equation
sin theta = 1/N
where N is the refractive index of water
Therefore theta = 48.7 degrees.
The radius of the circle through which light can escape the water surface, from a depth d below, is
R = d tan theta = 1.14 d
The diameter of the circle is 2R.
To find the diameter of the circle at the surface through which light emerges from the water, we can use Snell's Law.
Snell's Law states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the velocities of light in the two mediums, which in this case are air and water.
The formula for Snell's Law is:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
Where:
- n₁ and n₂ are the indices of refraction of the two mediums (in this case, air and water),
- θ₁ is the angle of incidence,
- θ₂ is the angle of refraction.
We can simplify Snell's Law for this problem by assuming the angle of incidence (θ₁) is 90 degrees, because the light is vertically entering the water.
So, the formula becomes:
n₁ = n₂ * sin(θ₂)
Given:
- n₁ = 1 (index of refraction of air),
- n₂ = 1.33 (index of refraction of water).
To find θ₂, we need to consider the concept of critical angle. When the angle of incidence exceeds the critical angle, the light would no longer refract and instead undergo total internal reflection.
The critical angle is given by:
sin(θ_c) = n₂ / n₁
Substituting the values, we have:
sin(θ_c) = 1.33 / 1
sin(θ_c) = 1.33
To find the angle of refraction (θ₂), we take the inverse sine (sin⁻¹) of the critical angle:
θ₂ = sin⁻¹(1.33)
Now that we have θ₂, we can determine the diameter of the circle at the surface through which light emerges from the water.
Let h be the height or depth of the water.
Let d be the diameter of the circle at the surface.
Using trigonometry, we can express h in terms of d and θ₂. Consider the right triangle formed by the depth of the water and the radius of the circle. The height (h) is the opposite side, and the radius (r) is the adjacent side.
We have:
tan(θ₂) = h / (d/2)
Rearranging the equation, we can express h in terms of d and θ₂:
h = (d/2) * tan(θ₂)
Given that h = 96.7 cm and θ₂ = sin⁻¹(1.33), we can substitute the known values into the equation:
96.7 = (d/2) * tan(sin⁻¹(1.33))
Now, we can solve for d by rearranging the equation and solving for d:
d = 2 * (96.7 / tan(sin⁻¹(1.33)))
Using a calculator, we can compute the value of d to find the diameter of the circle at the surface through which light emerges from the water.