Problem 89 - College Physics

Question

\(\bullet$$\bullet\) A thin layer of ice \((n=1.309)\) floats on the surface of water \((n=1.333)\) in a bucket. A ray of light from the bottom of the bucket travels upward through the water. (a) What is the largest angle with respect to the normal that the ray can make at the ice-water interface and still pass out into the air above the ice? (b) What is this angle after the ice melts?

Step-by-Step Solution

Answer

(a) 11.54°; (b) 48.75°.

Step 1: Understand the Concept

The problem requires understanding the concept of refraction and total internal reflection in optical mediums with different indices of refraction. We will use Snell's Law, which states that \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \), to find the critical angle where total internal reflection occurs.

Step 2: Determine Critical Angle at Ice-Water Interface

At the ice-water interface, the light ray passes from water \((n_2 = 1.333)\) to ice \((n_1 = 1.309)\). We need to find the critical angle \( \theta_c \) using Snell's Law with \( \theta_2 = 90^\circ \) (angle in air). Setting \( \sin \theta_2 = 1 \), we solve \( n_2 \sin \theta_c = n_1 \) at the critical angle.

Step 3: Calculate Critical Angle

Using the formula for the critical angle, \( \sin \theta_c = \frac{n_1}{n_2} = \frac{1.309}{1.333} \). Hence, \( \theta_c = \arcsin \left( \frac{1.309}{1.333} \right) \). Calculate this angle using a calculator.

Step 4: Solve for Air-Water Interface Without Ice

After the ice melts, the light travels directly from water \((n = 1.333)\) to air \((n = 1.0)\). Use \( \sin \theta_c = \frac{n_{air}}{n_{water}} = \frac{1}{1.333} \) to find the new critical angle. Calculate \( \theta_c = \arcsin \left( \frac{1}{1.333} \right) \).

Step 5: Conclude with Calculated Angles

With the values calculated: for ice-water interface, \( \theta_c \approx \arcsin(0.98) \) which is approximately 11.54°. After ice melts, \( \theta_c \approx 48.75° \).

Key Concepts

Critical Angle

The critical angle is a fundamental concept in optics, especially when dealing with light passing through different mediums. It is defined as the angle of incidence beyond which light is completely reflected back into the medium, instead of passing through the interface. This phenomenon occurs when light travels from a medium with a higher index of refraction to one with a lower index.

• For example, imagine light passing from water, with an index of refraction of 1.333, into air, having an index of 1.0. The critical angle is when the refraction turns into total internal reflection, and no light escapes into the air.
• To calculate the critical angle, you can use Snell's Law. Set the angle of refraction to 90°, meaning the refracted ray grazes along the interface. This simplifies Snell's Law to: \( n_1 \sin \theta_c = n_2 \sin 90^\circ \), giving the critical angle formula \( \sin \theta_c = \frac{n_2}{n_1} \).

Total Internal Reflection

Total internal reflection (TIR) is a fascinating occurrence in optics that happens when the light hits the boundary of two mediums at an angle greater than the critical angle. In this scenario, instead of refracting, the light is entirely reflected back into the original medium.

• This can only occur when light moves from a medium with a higher index of refraction toward a medium with a lower one. It's commonly observed in optical fibers, where light travels by bouncing off the fiber's sides without escaping, thanks to TIR.
• Take for instance the scenario in the exercise: the light traveling from water through ice before meeting the air. If the angle of incidence exceeds the calculated 11.54°, light will undergo total internal reflection instead of moving into the air.

Refraction in Optics

Refraction is the bending of light as it passes from one transparent medium to another. This occurs due to a change in the light's speed, which depends on the indices of refraction of both media it is transitioning between.

• When light enters a medium with a higher index of refraction (like going from air to water), it bends towards the normal line, slowing down in the process. Conversely, moving to a medium with a lower index causes the light to speed up and bend away from the normal.
• This principle is central in the given problem as the light moves from water to ice, altering its path. It changes again when the ice is absent, and the light moves directly into air, recalculating the critical angles using Snell's Law.

Indices of Refraction

Indices of refraction (n) are values that indicate how much light slows down when passing through a material. The index of refraction is a key element in predicting how light bends when moving between different substances.

• The higher the index, the more the light slows, and the more it bends. Water has an index of 1.333, ice is slightly less at 1.309, and air is 1.0, which is why their interfaces behave differently when light passes through.
• The exercise shows how the indices affect the critical angle and the conditions for total internal reflection as light interacts with different materials. Particularly, calculating these angles helps predict whether light will pass through or reflect back depending on the medium's indices.