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A transparent rod 30.0 cm long is cut flat at one end and rounded to a hemispherical surface of radius 10.0 cm at the other end. A small object is embedded within the rod along its axis and halfway between its ends, 15.0 cm from the flat end and 15.0 cm from the vertex of the curved end. When the rod is viewed from its flat end, the apparent depth of the object is 8.20 cm from the flat end. What is its apparent depth when the rod is viewed from its curved end?

Short Answer

Expert verified
The apparent depth from the curved end is approximately 10.02 cm.

Step by step solution

01

Understanding the problem

We know that for an object embedded in a medium, the apparent depth when viewed through different surfaces vary due to refraction. We're given the apparent depth from the flat end and need to determine it from the curved end.
02

Apparent Depth Formula

The apparent depth can be determined by the formula: \[ \text{Apparent Depth} = \frac{\text{Real Depth}}{n} \] where \( n \) is the refractive index of the material. Since we need to find apparent depth from a curved surface, a different equation will be needed in later steps.
03

Calculate the Refractive Index

Using the information provided, the apparent depth when viewed from the flat surface is 8.20 cm. The real depth is 15.0 cm. Using the refraction formula, we have: \( 8.20 = \frac{15.0}{n} \). Solving for \( n \), we get: \[ n = \frac{15.0}{8.20} = 1.829 \]
04

Use Lensmaker's Equation

To find the apparent depth when viewed from the curved end, apply the refraction formula for a curved surface: \[ \text{Apparent depth from curved surface } = \frac{s}{n - (n-1)\frac{R}{R+s}} \] where \( s = 15.0 \text{ cm} \) and \( R = 10.0 \text{ cm} \).
05

Plug into Formula

Insert values into the formula. Replace \( n \) with 1.829: \[ \text{Apparent Depth} = \frac{15.0}{1.829 - (1.829-1) \cdot \frac{10.0}{10.0+15.0}} \]
06

Simplify Equation

First calculate the correction factor: \( \frac{10.0}{25.0} = 0.4 \). Substituting values, the formula becomes: \[ \text{Apparent Depth} = \frac{15.0}{1.829 - 0.4 \times 0.829} \]
07

Final Calculations

Calculate the term: \(0.4 \times 0.829 = 0.3316\), which leads to: \[ \text{Apparent Depth} = \frac{15.0}{1.4974} \approx 10.02 \text{ cm} \]
08

Conclusion

When viewed from the curved end of the rod, the apparent depth of the object is approximately 10.02 cm from the vertex of the curved end.

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Key Concepts

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

Apparent Depth
Apparent depth is a fascinating optical phenomenon that occurs due to light refraction. Imagine peering into a clear pond, and noticing that the fish seem closer to the surface than they actually are. This is because as light passes from water to air, it bends, creating an optical illusion of depth.

Apparent depth can be determined practically using the formula:
  • Apparent Depth = \( \frac{\text{Real Depth}}{n} \)
where \( n \) is the refractive index of the medium. This formula gives us a glimpse into how much closer an object appears when it is submerged in a material with a refractive index different from its surroundings. It's essential in fields like underwater photography and aquariums, where the perception of distance is altered due to the medium's properties.
Refractive Index
The refractive index plays a crucial role in determining how much light bends as it enters a new medium. This dimensionless number essentially hints at how much the speed of light is reduced inside the material.

Refractive index \( n \) can be calculated using:
  • \( n = \frac{c}{v} \)
where \( c \) is the speed of light in a vacuum and \( v \) is the speed of light in the medium.

The higher the refractive index, the more the light bends. In our exercise, the refractive index of the rod material was calculated to be approximately 1.829, based on the change in apparent depth when viewed from its flat end.
Lensmaker's Equation
The lensmaker’s equation is a powerful tool in optics, especially useful for designing lenses with desired focusing properties. It gives the relationship between a lens's shape, its refractive index, and its focal length.

While the full lensmaker's equation is often used for lenses, in the case of a single curved surface like the hemispherical end of our rod, a simplified refraction formula for curved surfaces can be employed:
  • Apparent depth from the curved surface is given by:
  • \( \frac{s}{n - (n-1)\frac{R}{R+s}} \)
where:
  • \( s \) is the object's distance from the surface
  • \( R \) is the radius of curvature
  • \( n \) is the refractive index
This equation was crucial in determining how the position of the object appears differently when looking from the rod's curved surface.
Curved Surface Refraction
Curved surface refraction deals with how light behaves as it passes through a surface that is not flat. This is particularly relevant in the world of lenses and rounded objects, like our transparent rod.

When light encounters a curved boundary, it doesn't just change speed and bend; it can also converge or diverge depending on the curvature. Using the adjusted formula for apparent depth considering the curved surface helped us solve the exercise's problem by properly accounting for these effects.

By understanding the way refraction operates through a hemispherical surface, students can better appreciate practical applications like lens crafting and optical illusions.

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Most popular questions from this chapter

The smallest object we can resolve with our eye is limited by the size of the light receptor cells in the retina. In order for us to distinguish any detail in an object, its image cannot be any smaller than a single retinal cell. Although the size depends on the type of cell (rod or cone), a diameter of a few microns (\(\mu\)m) is typical near the center of the eye. We shall model the eye as a sphere 2.50 cm in diameter with a single thin lens at the front and the retina at the rear, with light receptor cells 5.0 \(\mu\)m in diameter. (a) What is the smallest object you can resolve at a near point of 25 cm? (b) What angle is subtended by this object at the eye? Express your answer in units of minutes (1\(^\circ\) = 60 min), and compare it with the typical experimental value of about 1.0 min. (\(Note\): There are other limitations, such as the bending of light as it passes through the pupil, but we shall ignore them here.)

A thin lens with a focal length of 6.00 cm is used as a simple magnifier. (a) What angular magnification is obtainable with the lens if the object is at the focal point? (b) When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at the near point, 25.0 cm from the eye, and that the lens is very close to the eye.

An object to the left of a lens is imaged by the lens on a screen 30.0 cm to the right of the lens. When the lens is moved 4.00 cm to the right, the screen must be moved 4.00 cm to the left to refocus the image. Determine the focal length of the lens.

The radii of curvature of the surfaces of a thin converging meniscus lens are \(R_1\) = +12.0 cm and \(R_2\) = +28.0 cm. The index of refraction is 1.60. (a) Compute the position and size of the image of an object in the form of an arrow 5.00 mm tall, perpendicular to the lens axis, 45.0 cm to the left of the lens. (b) A second converging lens with the same focal length is placed 3.15 m to the right of the first. Find the position and size of the final image. Is the final image erect or inverted with respect to the original object? (c) Repeat part (b) except with the second lens 45.0 cm to the right of the first.

You wish to project the image of a slide on a screen 9.00 m from the lens of a slide projector. (a) If the slide is placed 15.0 cm from the lens, what focal length lens is required? (b) If the dimensions of the picture on a 35-mm color slide are 24 mm \(\times\) 36 mm, what is the minimum size of the projector screen required to accommodate the image?

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