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Chapter 34: Problem 95

One end of a long glass rod is ground to a convex hemispherical shape. This glass has an index of refraction of 1.55 . When a small leaf is placed 20.0 \(\mathrm{cm}\) in front of the center of the hemisphere along the optic axis, an image is formed inside the glass 9.12 \(\mathrm{cm}\) from the spherical surface. Where would the image be formed if the glass were now immersed in water (refractive index 1.33 ) but nothing else were changed?

Short Answer

Expert verified
In water, the image is formed 10.03 cm from the spherical surface.

Step by step solution

01

Identify the Lensmaker's Equation

The lensmaker's equation for a spherical refracting surface is given by \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \), where \( n_1 \) and \( n_2 \) are the refractive indices of the media from which light is incident and into which it refracts, \( u \) is the object distance, \( v \) is the image distance, and \( R \) is the radius of curvature of the surface.
02

Calculate Radius of Curvature in Air

Given that \( n_1 = 1 \) (air), \( n_2 = 1.55 \) (glass), \( u = -20 \) cm, and \( v = 9.12 \) cm, we substitute these values into the lensmaker's formula: \( \frac{1.55}{9.12} - \frac{1}{-20} = \frac{1.55 - 1}{R} \). Solve for \( R \) to find the radius of curvature.
03

Solve for Radius of Curvature

Rearrange and solve the equation: \( \frac{1.55}{9.12} + \frac{1}{20} = \frac{0.55}{R} \). Calculate \( \frac{1.55}{9.12} \approx 0.170 \, \text{ and } \, \frac{1}{20} = 0.05 \), which gives \( 0.220 = \frac{0.55}{R} \). Thus, \( R = \frac{0.55}{0.220} \approx 2.5 \) cm.
04

Re-calculate Image Position in Water

Now that the glass is immersed in water, we have: \( n_1 = 1.33 \) (water), \( n_2 = 1.55 \) (glass), and the previously calculated \( R = 2.5 \) cm. Use the lensmaker's formula again: \( \frac{1.55}{v} - \frac{1.33}{-20} = \frac{1.55 - 1.33}{2.5} \).Solve this to find the new image distance \( v \).
05

Solve for New Image Distance

With \( \frac{0.22}{2.5} = 0.088 \), the equation becomes: \( \frac{1.55}{v} = 0.088 + \frac{1.33}{20} \approx 0.088 + 0.0665 = 0.1545 \).Solving gives \( \frac{1.55}{0.1545} = v \approx 10.03 \) cm.

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Key Concepts

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

Refractive Index
The refractive index is a crucial concept in understanding how light behaves when traveling between different media. It's a measure of how much the speed of light is reduced inside a material compared to in a vacuum. The refractive index (\( n \)) can greatly affect how lenses focus light.
  • When light enters a medium with a higher refractive index, it slows down and bends towards the normal line - the imaginary line perpendicular to the surface.
  • The change in speed and angle is described by Snell's Law, linking the angles and refractive indices of the two media.
  • This property is what allows lenses to focus light to form images, making it pivotal in lens design and analysis.
In our problem, glass has a refractive index of 1.55. This means light travels slower in the glass than it would in water, which has a refractive index of 1.33. The difference in these indices directly affects where images are formed when light passes from one medium to another.
Radius of Curvature
The radius of curvature (\( R \)) is a measure of the distance from the center of a lens or spherical mirror to its surface. It is essential in determining how a lens will refract or bend light.
  • In the context of the lensmaker's equation, the radius of curvature helps ascertain the focusing ability of a lens with a spherical surface.
  • A larger radius means a flatter surface, while a smaller radius results in a more curved surface.
  • Curvature influences how convergently or divergently a lens bends light rays.
In solving our exercise, we calculated the radius of curvature to be 2.5 cm using the lensmaker's equation. This value tells us how curved the glass rod's hemispherical end is, which is crucial for determining the focal point and image position. Adjustments in this parameter can dramatically alter how and where an image is formed.
Optic Axis
The optic axis is a fundamental concept in optics, often referred to in discussions involving lenses and optical systems. It is an imaginary line that defines the path along which light travels through a lens or optical system.
  • It generally passes through the geometric center of the lens.
  • In spherical lenses, the optic axis represents the line along which symmetry of the lens is intact, ensuring predictable light refraction.
  • The optic axis is vital for accurately aligning optical elements in devices like cameras and telescopes.
In the context of our problem, the optic axis is where we place the object (the leaf) relative to the hemispherical glass. Correct alignment ensures that we can predictably calculate where an image will form and illustrate effective use of the lensmaker's equation.

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Most popular questions from this chapter

A converging lens forms an image of an 8.00 -mm-tall real object. The image is 12.0 \(\mathrm{cm}\) to the left of the lens, 3.40 \(\mathrm{cm}\) tall, and erect. What is the focal length of the lens? Where is the object located?

A certain microscope is provided with objectives that have focal lengths of \(16 \mathrm{mm}, 4 \mathrm{mm},\) and 1.9 \(\mathrm{mm}\) and with eye-pieces that have angular magnifications of \(5 \times\) and \(10 \times .\) Each objective forms an image 120 \(\mathrm{mm}\) beyond its second focal point. Determine (a) the largest overall angular magnification obtainable and (b) the smallest overall angular magnification obtainable.

The radii of curvature of the surfaces of a thin converging meniscus lens are \(R_{1}=+12.0 \mathrm{cm}\) and \(R_{2}=+28.0 \mathrm{cm} .\) 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 \(\mathrm{mm}\) tall, perpendicular to the lens axis, 45.0 \(\mathrm{cm}\) to the left of the lens. (b) A second converging lens with the same focal length is placed 3.15 \(\mathrm{m}\) 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 \(\mathrm{cm}\) to the right of the first.

An object is 24.0 \(\mathrm{cm}\) from the center of a silvered spherical glass Christmas tree ornament 6.00 \(\mathrm{cm}\) in diameter. What are the position and magnification of its image?

Recall that the intensity of light reaching film in a camera is proportional to the effective area of the lens. Camera A has a lens with an aperture diameter of 8.00 \(\mathrm{mm} .\) It photographs an object using the correct exposure time of \(\frac{1}{30} \mathrm{s}\) . What exposure time should be used with camera \(\mathrm{B}\) in photographing the same object with the same film if this camera has a lens with an aperture diameter of 23.1 \(\mathrm{mm}\) ?
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