Question

A vessel of depth \(\mathrm{t}\) is half filled with oil of refractive index \(\mathrm{n}_{1}\) and the other half is filled with water (refractive index \(\mathrm{n}_{2}\) ). The apparent depth of the vessel when viewed from above is \(\ldots \ldots\) (A) \(\left[\left\\{2 t\left(n_{1}-\mathrm{n}_{2}\right)\right\\} /\left(\mathrm{n}_{1} \mathrm{n}_{2}\right)\right]\) (B) \(\left[\left\\{2 \mathrm{t}\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)\right\\} /\left(\mathrm{n}_{1} \mathrm{n}_{2}\right)\right]\) (C) \(\left[\left\\{t\left(n_{1}-n_{2}\right)\right\\} /\left(2 n_{1} \mathrm{n}_{2}\right)\right]\) (D) \(\left[\left\\{t\left(n_{1}+n_{2}\right)\right\\} /\left(2 n_{1} \mathrm{n}_{2}\right)\right]\)

Short answer

The apparent depth of the vessel when viewed from above is \(\boxed{\text{(D)}\ \left[\frac{t\left(n_{1}+n_{2}\right)}{2 n_{1}n_{2}}\right]} \).

Step-by-step solution

Calculate the actual depths of oil and water

As the vessel is half-filled with oil and half-filled with water, the actual depths of oil and water in the vessel can be calculated as follows: Actual depth of oil: \(t_{1} = \frac{t}{2}\) Actual depth of water: \(t_{2} = \frac{t}{2}\)

Calculate apparent depths of oil and water

The formula for calculating the apparent depth, given the actual depth and refractive index, is: Apparent depth = \(\frac{\text{Actual Depth}}{\text{Refractive Index}}\) Using this formula, we'll calculate the apparent depths of oil and water: For oil: Apparent depth \(p_{1} = \frac{t_{1}}{n_{1}} = \frac{\frac{t}{2}}{n_{1}}\) For water: Apparent depth \(p_{2} = \frac{t_{2}}{n_{2}} = \frac{\frac{t}{2}}{n_{2}}\)

Calculate the total apparent depth of the vessel

To find the total apparent depth of the vessel, we will add up the apparent depths of oil and water: Total apparent depth \(p_{t} = p_{1} + p_{2}\) Plugging in the values we found in Step 2: \(p_{t} = \frac{\frac{t}{2}}{n_{1}} + \frac{\frac{t}{2}}{n_{2}}\) By finding a common denominator, we can simplify this expression further: \(p_{t} = \frac{t(n_{1} + n_{2})}{2n_{1}n_{2}}\) Comparing this expression with the given options, we find that the apparent depth of the vessel when viewed from above is: \[\boxed{\text{(D)}\ \left[\frac{t\left(n_{1}+n_{2}\right)}{2 n_{1}n_{2}}\right]} \]

Key concepts

Refractive Index

The refractive index is a measure of how much light bends, or refracts, as it passes from one medium to another. It provides insight into how a material will change the direction of light passing through it. Defined mathematically, the refractive index of a material is the ratio of the speed of light in a vacuum to the speed of light in the medium.

Symbolically, it's denoted as:

• Refractive Index () = \\(\frac{c}{v}\), where \\(c\) is the speed of light in a vacuum, and \\(v\) is the speed of light in the medium.

This concept is essential in understanding how light behaves when moving between different substances like air, water, or oil. Each substance has its specific refractive index, leading to different bending behaviors. Understanding this concept is crucial for solving optics problems like the one in our exercise.

Apparent Depth Calculation

Apparent depth is the perceived depth of an object submerged in a transparent medium, viewed through another medium. Due to the refractive index difference, objects submerged in a medium appear shallower than their actual depth when viewed from above.

The formula for apparent depth is:

• Apparent Depth = \(\frac{\text{Actual Depth}}{\text{Refractive Index}}\)

In our example, the vessel is half-filled with oil and half with water. To calculate the total apparent depth, we find the apparent depths of both liquids separately using their refractive indices and then add them together. This is a straightforward yet powerful way to understand how light behavior can alter our perception of depth in real-world scenarios.

Optics Problem

Optics problems involve understanding how light interacts with different media. These problems require knowledge of concepts like refraction, reflection, and diffraction, each explained by specific laws such as Snell's Law.

They often involve:

• Calculating how light bends as it enters a new medium, depending on the refractive index.
• Understanding how this bending affects the perceived position or depth of objects.

By solving optics problems, one can predict and explain various phenomena such as why a straw appears bent in a glass of water or why submerged objects seem closer than they actually are. The exercise we worked through is a classic example of an optics problem that challenges our understanding of real versus perceived depth due to light refraction.

Light Refraction in Liquids

Light refraction occurs when light passes from one medium to another, changing speed and bending in the process. This bending is most noticeable when light enters or exits materials like water or glass, which have refractive indices significantly different from air.

In liquids, light refraction plays a crucial role in:

• Determining how images are distorted or magnified.
• Affecting visibility and apparent positions of objects under water.

For example, in our exercise, the light passing from the air through oil and water alters its path. This makes the bottom of the vessel appear at a different depth than it actually is. Such interactions can be visually striking and are essential in fields like optical engineering and underwater photography. By understanding refraction in liquids, we can better design lenses, lighting, and imaging systems to accommodate or utilize this behavior.