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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.

Short Answer

Expert verified
The final image is inverted and taller for both positions, with significant changes in size due to distance differences between the lenses.

Step by step solution

01

Calculate Focal Length of the Meniscus Lens

We use the lens maker's formula to find the focal length:\[ \frac{1}{f} = (n-1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]Given \( R_1 = +12.0 \, \text{cm} \), \( R_2 = +28.0 \, \text{cm} \), and \( n = 1.60 \), substitute the values:\[ \frac{1}{f} = (1.60 - 1) \left(\frac{1}{12.0} - \frac{1}{28.0}\right) \]Calculate to find \( f \).
02

Calculate Image Position with Lens Equation (Part a)

Use the lens equation:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Given \( d_o = 45.0 \, \text{cm} \) and \( f \) from Step 1, solve for \( d_i \).Rearrange:\[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \]Plug in the values to find the image distance \( d_i \).
03

Determine the Image Magnification and Size (Part a)

Calculate magnification \( m \) using:\[ m = -\frac{d_i}{d_o} \]Find the height of the image \( h_i \) using:\[ h_i = m \times h_o \]Where \( h_o = 5.00 \, \text{mm} \). Substitute to find \( h_i \).
04

Find Image Position for Systems with Second Lens at 3.15 m (Part b)

The system consists of two lenses; find the image from the first lens and use that as the object for the second lens.\( d'_o \), the distance of the object for the second lens, is:\[ d'_o = 315 \, \text{cm} - d_i \]Use the lens equation for the second lens again to find new \( d'_i \).
05

Calculate Final Image Size and Orientation for 3.15 m Scenario (Part b)

Calculate the magnification of the second lens using:\[ m' = -\frac{d'_i}{d'_o} \]Calculate the total magnification:\[ M = m \times m' \]Find the final image height:\[ h'_i = M \times h_o \]Determine orientation based on the total magnification.
06

Repeat Steps 4 and 5 for Lens at 45.0 cm (Part c)

Change \( d'_o \) to 45.0 cm from the first lens's \( d_i \). Use the same calculations as in Steps 4 and 5 to find the position, size, and orientation of the image.

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Key Concepts

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

Lens Maker's Formula
The lens maker's formula is a fundamental equation for determining the focal length of a lens, especially useful for lenses with different curvatures on each side. It is expressed as:\[ \frac{1}{f} = (n-1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]where:
  • \(f\) is the focal length of the lens.
  • \(n\) is the refractive index of the lens material.
  • \(R_1\) and \(R_2\) are the radii of curvature of the lens surfaces.
If both surfaces of the lens are curved, the formula shows how the difference in their curvatures, along with the material's refractive index, determines the lens's ability to converge or diverge light. Understanding this formula is essential for solving problems involving lenses with complex shapes, such as the meniscus lens in this exercise.
Thin Lens Equation
The thin lens equation is vital for determining the position of an image formed by a lens. Mathematically, it is given by:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]where:
  • \(d_o\) is the distance from the object to the lens.
  • \(d_i\) is the distance from the image to the lens.
  • \(f\) is the focal length, calculated using the lens maker's formula.
By rearranging the equation, you can solve for the unknown distance, providing you with either \(d_o\) or \(d_i\) depending on what's required. This equation is crucial for pinpointing where an image will form in relation to the lens, which helps in understanding complex systems like multiple lens setups, as seen in this step-by-step solution.
Image Magnification
Image magnification provides insight into how an object's size changes when viewed through a lens. It's calculated using the formula:\[ m = -\frac{d_i}{d_o} \]where:
  • \(m\) is the magnification.
  • \(d_i\) is the image distance.
  • \(d_o\) is the object distance.
The negative sign indicates the image's orientation. A positive magnification value means the image is upright compared to the object, while a negative value indicates it is inverted. Additionally, calculating the image height using magnification helps determine how large the image appears compared to the actual object height, using this formula:\[ h_i = m \times h_o \]where \(h_o\) is the object's real height.Understanding magnification is essential when dealing with practical applications, such as photography or optical instrumentation, to predict the size and orientation of an image.
Refractive Index
The refractive index, represented by \(n\), characterizes how much a medium bends light. It is defined as the ratio of the speed of light in vacuum to the speed of light in the medium:\[ n = \frac{c}{v} \]where:
  • \(c\) is the speed of light in a vacuum.
  • \(v\) is the speed of light in the medium.
In geometrical optics, the refractive index is crucial because it influences the bending of light as it passes through different materials, a process captured by Snell's Law. For lenses, the refractive index significantly impacts how a lens focuses light, as seen in the lens maker’s formula, where the value of \(n\) subtracts 1 to adjust for how the material of the lens impacts focal length. A higher refractive index means greater bending of light, which is why denser materials, like glass or diamond, are used for lenses.

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Most popular questions from this chapter

A solid glass hemisphere of radius 12.0 \(\mathrm{cm}\) and index of refraction \(n=1.50\) is placed with its flat face downward on a table. A parallel beam of light with a circular cross section 3.80 \(\mathrm{mm}\) in diameter travels straight down and enters the hemisphere at the center of its curved surface. (a) What is the diameter of the circle of light formed on the table? (b) How does your result depend on the radius of the hemisphere?

(a) For a lens with focal length \(f,\) find the smallest distance possible between the object and its real image. (b) Graph the distance between the object and the real image as a function of the distance of the object from the lens. Does your graph agree with the result you found in part (a)?

A telescope is constructed from two lenses with focal lengths of 95.0 \(\mathrm{cm}\) and \(15.0 \mathrm{cm},\) the \(95.0-\mathrm{cm}\) lens being used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnification for the telescope. (b) Find the height of the image formed by the objective of a building 60.0 \(\mathrm{m}\) tall, 3.00 \(\mathrm{km}\) away. (c) What is the angular size of the final image as viewed by an eye very close to the eyepiece?

A converging lens with a focal length of 90.0 \(\mathrm{cm}\) forms an image of a 3.20 -cm-tall real object that is to the left of the lens. The image is 4.50 \(\mathrm{cm}\) tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?

Corrective Lenses. Determine the power of the corrective contact lenses required by (a) a hyperopic eye whose near point is at 60.0 \(\mathrm{cm}\) and \((b)\) a myopic eye whose far point is at \(60.0 \mathrm{cm} .\)

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