A cup with refractive index n2 = 1.8 has outer radius b = 6.0 cm and inner radius a = 4.0 cm. It is filled with a liquid with refractive index n3 = 1.3. In the surrounding medium with n1 = 1.0,a ray of light travels along a trajectory that is at a perpendicular distance y from the center of the cup. The ray enters the cup and, reaching the liquid interface at the critical angle, undergoes total internal reflection. Determine the distance y
To determine the distance y at which total internal reflection occurs, we need to apply Snell's Law and the concept of critical angle.
Let's start by considering the light ray as it travels from the surrounding medium (n1 = 1.0) into the cup (n2 = 1.8). Since total internal reflection occurs at the liquid interface (n3 = 1.3), we know that the angle of incidence at this interface must be equal to the critical angle.
The critical angle (θc) is the angle at which light incident on the interface will be reflected back into the medium. It can be calculated using the formula:
θc = sin^(-1)(n2/n3)
Substituting in the given values, we have:
θc = sin^(-1)(1.8/1.3) ≈ 57.24°
Next, we need to find the angle of incidence at the liquid interface. Since the ray is traveling along a trajectory at a distance y from the center of the cup, the angle of incidence will be given by:
θi = tan^(-1)(y/r)
where r is the distance from the center of the cup to the point where the light ray enters the cup. In this case, r is the average of the inner and outer radii:
r = (a + b)/2 = (4.0 cm + 6.0 cm)/2 = 5.0 cm
Finally, to determine the distance y at which total internal reflection occurs, we equate the angle of incidence (θi) to the critical angle (θc):
θi = θc
tan^(-1)(y/5.0 cm) = 57.24°
To solve this equation for y, we can take the tangent of both sides:
y/5.0 cm = tan(57.24°)
y = 5.0 cm * tan(57.24°)
Using a calculator, we find:
y ≈ 7.07 cm
Therefore, the distance y at which total internal reflection occurs is approximately 7.07 cm.