Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A light bulb is 3.00 m from a wall. You are to use a concave mirror to project an image of the bulb on the wall, with the image 3.50 times the size of the object. How far should the mirror be from the wall? What should its radius of curvature be?

A convex spherical mirror with a focal length of magnitude 24.0 cm is placed 20.0 cm to the left of a plane mirror. An object 0.250 cm tall is placed midway between the surface of the plane mirror and the vertex of the spherical mirror. The spherical mirror forms multiple images of the object. Where are the two images of the object formed by the spherical mirror that are closest to the spherical mirror, and how tall is each image?

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.

Three thin lenses, each with a focal length of 40.0 cm, are aligned on a common axis; adjacent lenses are separated by 52.0 cm. Find the position of the image of a small object on the axis, 80.0 cm to the left of the first lens.

The cornea of the eye has a radius of curvature of approximately 0.50 cm, and the aqueous humor behind it has an index of refraction of 1.35. The thickness of the cornea itself is small enough that we shall neglect it. The depth of a typical human eye is around 25 mm. (a) What would have to be the radius of curvature of the cornea so that it alone would focus the image of a distant mountain on the retina, which is at the back of the eye opposite the cornea? (b) If the cornea focused the mountain correctly on the retina as described in part (a), would it also focus the text from a computer screen on the retina if that screen were 25 cm in front of the eye? If not, where would it focus that text: in front of or behind the retina? (c) Given that the cornea has a radius of curvature of about 5.0 mm, where does it actually focus the mountain? Is this in front of or behind the retina? Does this help you see why the eye needs help from a lens to complete the task of focusing?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free