Question

A layer of benzene (\(n\) = 1.502) that is 4.20 cm deep floats on water (\(n\) = 1.332) that is 5.70 cm deep. What is the apparent distance from the upper benzene surface to the bottom of the water when you view these layers at normal incidence?

Short answer

The apparent distance is 5.65 cm.

Step-by-step solution

Calculate Real Depth

The real depth is the actual thickness of the layers stacked. Here we need to add the depth of the benzene and water layers: 4.20 cm (benzene) + 5.70 cm (water) = 9.90 cm. So, the real depth of the entire fluid layer is 9.90 cm.

Determine Apparent Depth Equation

When viewing through a substance, the apparent depth can be calculated using the formula: \[ d' = \frac{d}{n} \]where \(d'\) is the apparent depth, \(d\) is the actual depth, and \(n\) is the refractive index of the medium. This equation is used individually for each layer.

Calculate Apparent Depth of Benzene

Using the equation for apparent depth, calculate it for the benzene layer:\[ d'_\text{benzene} = \frac{4.20 \text{ cm}}{1.502} \approx 2.80 \text{ cm} \].The apparent depth of the benzene layer is 2.80 cm.

Calculate Apparent Depth of Water Seen Through Benzene

For the water layer, which is viewed through benzene, the apparent depth calculation needs two steps: first calculate the apparent depth as viewed directly, then correct for the benzene refraction:1. Apparent depth of the water alone: \\[ d'_\text{water-actual} = \frac{5.70 \text{ cm}}{1.332} \approx 4.28 \text{ cm} \].2. Then, since we view the water through benzene, adjust for benzene: \[ d'_\text{water-seen through benzene} = \frac{4.28 \text{ cm}}{1.502} \approx 2.85 \text{ cm} \].The apparent depth of the water seen through benzene is 2.85 cm.

Sum Apparent Depths

To find the total apparent distance from the upper benzene surface to the bottom of the water, sum the individual apparent depths:\[ d'_{\text{total}} = d'_\text{benzene} + d'_\text{water-seen through benzene} \approx 2.80 \text{ cm} + 2.85 \text{ cm} = 5.65 \text{ cm}. \]The total apparent distance is 5.65 cm.

Key concepts

Refractive Index

The refractive index is a measure of how much light bends, or refracts, as it enters a medium from another. Every material has a refractive index, denoted by the symbol \( n \). It tells us how much slower light travels in the medium compared to the speed of light in a vacuum.

• If \( n = 1.502 \), it means light travels 1.502 times slower in that medium compared to a vacuum.
• Materials with a higher refractive index bend light more significantly.
• For example, benzene has a refractive index of \( 1.502 \) while water's is \( 1.332 \).

Understanding this helps us comprehend why things appear shallower in water or other materials—a phenomenon crucial in solving our apparent depth problem.

Snell's Law

Snell's Law describes how light refracts when moving from one medium to another. It states that the ratio of the sines of the angles of incidence (angle where the light hits a surface) and refraction (angle where the light travels in the new medium) is constant, and equal to the ratio of the refractive indices of the two media.

For instance, if light passes from benzene to water:

• The angle of incidence \( \theta_1 \) in benzene and the angle of refraction \( \theta_2 \) in water follow the relation: \[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]
• Where \( n_1 \) is the refractive index of benzene and \( n_2 \) is that of water.

Although in the exercise the incidence is normal (perpendicular) thus no refraction angle change, understanding Snell's Law helps see how light behavior affects apparent depth.

Physics Problem Solving

In physics problem-solving, breaking down the problem into manageable steps is key. Whether determining apparent depth or any other challenge, here's how you can tackle it effectively:

• Identify and list all given data, like depths and refractive indices for this problem.
• Understand the underlying physics principles—in this instance, refractive indices, and their relation to light bending.
• Use appropriate formulas, such as apparent depth \( d' = \frac{d}{n} \), to address each part of the problem individually.
• Combine results to get the final answer, summing the layers' apparent depths like in our solution.

Taking a systematic approach simplifies even complex physics problems, making them less daunting and easier to solve.