/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 A small object is placed at the ... [FREE SOLUTION] | 91Ó°ÊÓ

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A small object is placed at the centre of the bottom of a cylindrical vessel of radius \(3 \mathrm{~cm}\) and height \(4 \mathrm{~cm}\) filled completely with water. Consider the ray leaving the vessel through a corner. Suppose this ray and the ray along the axis of the vessel are used to trace the image. Find the apparent depth of the image and the ratio of real depth to the apparent depth under the assumptions taken. Refractive index of water \(=1 \cdot 33\).

Short Answer

Expert verified
The apparent depth is approximately \(3.01 \mathrm{~cm}\), with a ratio of real depth to apparent depth of about 1.33.

Step by step solution

01

Gather the Information

Identify all the given information for the problem: cylindrical vessel with a radius of \(3 \mathrm{~cm}\) and height of \(4 \mathrm{~cm}\), refractive index of water \(n = 1.33\). The goal is to find the apparent depth of the object and the ratio of real depth to apparent depth.
02

Understand Refraction Concept

The apparent depth of an object submerged in a medium is given by the formula \( \text{apparent depth} = \frac{\text{real depth}}{n} \), where \(n\) is the refractive index of the medium (1.33 for water in this case).
03

Calculate Real Depth

The real depth of the object is the actual distance from the object to the surface of the water, which is given as the height of the cylinder, i.e., \( 4 \mathrm{~cm} \).
04

Compute Apparent Depth

Using the formula for apparent depth, we calculate:\[ \text{apparent depth} = \frac{4 \mathrm{~cm}}{1.33} \approx 3.01 \mathrm{~cm} \]
05

Determine the Ratio of Depths

Now, calculate the ratio of real depth to apparent depth:\[ \frac{\text{real depth}}{\text{apparent depth}} = \frac{4 \mathrm{~cm}}{3.01 \mathrm{~cm}} \approx 1.33 \] This confirms the relationship based on refractive index.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Refractive Index
Refractive index is a fundamental concept in optics that measures how much light is bent, or refracted, when it enters a different medium. It is a dimensionless number indicating the speed of light in a vacuum divided by the speed of light in the given medium.
For example, the refractive index of water is 1.33, which means light travels 1.33 times slower in water than in a vacuum. This property is crucial in understanding phenomena like apparent depth.
When light travels from water to air, it speeds up, causing the light path to bend. This bending makes objects submerged in water appear closer to the surface than they actually are. Hence, the concept of apparent depth arises, which is closely linked to the refractive index of the medium the light is traveling through.
Cylindrical Vessel
A cylindrical vessel, in this context, serves as a container filled with water. It's important to consider its dimensions, such as the radius and height, especially when calculating the real and apparent depths.
The given problem specifies a radius of 3 cm and a height of 4 cm. These dimensions define the path of light rays from the submerged object to the water's surface. In a cylinder, light can travel at angles depending on where the light exits, not just directly upward. Therefore, understanding the shape and orientation of the vessel helps in tracing the path of the rays, which is essential to finding the apparent depth.
In practical scenarios, if the vessel's geometry changes, it could affect the perceived image and the calculations needed for finding optical properties like apparent depth.
Real Depth
Real depth refers to the actual distance from a submerged object to the surface of the medium it is in. In the exercise, the real depth is given by the height of the cylindrical vessel, which is 4 cm. This is the vertical distance light travels from the object to the surface of the water.
Knowing the real depth is crucial for calculations involving optical phenomena, such as apparent depth. Apparent depth calculations use the ratio of real depth to refractive index, linking it closely to the overall understanding of refraction.
In a simplified optical system involving a single medium (like water), real depth is straightforward to measure physically but is altered perceptually by the refractive properties of the medium.
Refraction Concept
Refraction is the bending of light as it passes from one medium to another of different densities, such as from water to air. This change in speed results in a change in direction, governed by Snell's law.
In the exercise, refraction explains why the object's apparent depth inside the water appears different from its real depth. To calculate apparent depth, the formula used is \( \text{apparent depth} = \frac{\text{real depth}}{n} \), where \( n \) is the refractive index of the water. The apparent depth is less than the real depth due to the light bending upward as it exits the water.
Understanding refraction is fundamental for solving optical problems, as it helps explain how light interactions can alter perceptions and measurements in different mediums.

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Most popular questions from this chapter

A man uses a concave mirror for shaving. He keeps his face at a distance of \(25 \mathrm{~cm}\) from the mirror and gets an image which is \(1 \cdot 4\) times enlarged. Find the focal length of the mirror.

A small block of mass \(m\) and a concave mirror of radius \(R\) fitted with a stand lie on a smooth horizontal table with a separation \(d\) between them. The mirror together with its stand has a mass \(m .\) The block is pushed at \(t=0\) towards the mirror so that it starts moving towards the mirror at a constant speed \(V\) and collides with it. The collision is perfectly elastic. Find the velocity of the image (a) at a time \(td / V\).

A thin lens made of a material of refractive index \(\mu_{2}\) has a medium of refractive index \(\mu_{1}\) on one side and a medium of refractive index \(\mu_{3}\) on the other side. The lens is biconvex and the two radii of curvature have equal magnitude \(R\). A beam of light travelling parallel to the principal axis is incident on the lens. Where will the image be formed if the beam is incident from (a) the medium \(\mu_{1}\) and (b) from the medium \(\mu_{3}\) ?

A ball is kept at a height \(h\) above the surface of a heavy transparent sphere made of a material of refractive index \(\mu\). The radius of the sphere is \(R\). At \(t=0\), the ball is dropped to fall normally on the sphere. Find the speed of the image formed as a function of time for \(t<\sqrt{\frac{2 h}{g}}\). Consider only the image by a single refraction.

A particle executes a simple harmonic motion of amplitude \(1 \cdot 0 \mathrm{~cm}\) along the principal axis of a convex lens of focal length \(12 \mathrm{~cm} .\) The mean position of oscillation is at \(20 \mathrm{~cm}\) from the lens. Find the amplitude of oscillation of the image of the particle.

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