The index of refraction of a transparent liquid
(similar to water but with a different index of refraction) is 1.39. A flashlight held under the
transparent liquid shines out of the transparent liquid in a swimming pool. This beam of light exiting the surface of the transparent liquid makes an angle of �a = 39 degrees with respect
to the vertical.
At what angle (with respect to the vertical)is the flashlight being held under transparent liquid?
Answer in units of degrees.
The flashlight is slowly turned away from the vertical direction.
At what angle will the beam no longer be visible above the surface of the pool? Answer in units of degrees.
To find the angle at which the flashlight is being held under the transparent liquid (angle of incidence), we can use Snell's law, which relates the angle of incidence and the refractive indices of the two mediums.
Snell's law states:
n1 * sin(theta1) = n2 * sin(theta2)
Where:
n1 = refractive index of the medium the light is coming from (air)
theta1 = angle of incidence
n2 = refractive index of the medium the light is entering (transparent liquid)
theta2 = angle of refraction
Given:
n1 = 1 (approximation for air)
n2 = 1.39
theta2 = 39 degrees
Rearranging Snell's law, we can solve for theta1:
theta1 = asin((n2/n1) * sin(theta2))
Substituting the known values:
theta1 = asin((1.39/1) * sin(39))
Calculating this, we find theta1 ≈ 55.4 degrees.
Therefore, the flashlight is being held at an angle of approximately 55.4 degrees with respect to the vertical.
For the second part of the question, regarding the angle at which the beam will no longer be visible above the surface of the pool, we need to consider the concept of Total Internal Reflection.
Total Internal Reflection occurs when the angle of incidence is greater than the critical angle, which is the angle at which light rays no longer refract but instead reflect back inside the medium.
The critical angle can be calculated using the equation:
theta_c = asin(n2/n1)
Given:
n1 = 1 (approximation for air)
n2 = 1.39
theta_c = asin(1.39/1)
Calculating this, we find theta_c ≈ 52.8 degrees.
Therefore, when the flashlight is turned beyond an angle of approximately 52.8 degrees away from the vertical, the beam of light will no longer be visible above the surface of the pool.