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Chapter 34: Problem 25

An insect 3.75 mm tall is placed 22.5 \(\mathrm{cm}\) to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude \(13.0 \mathrm{cm},\) and the index of refraction of the lens material is 1.70. (a) Calculate the location and size of the image this lens forms of the insect. Is it real or virtual? Erect or inverted? (b) Repeat part (a) if the lens is reversed.

Short Answer

Expert verified
(a) The image is virtual, 60.00 cm right, 10.0 mm tall, upright. (b) The image is real, 90.00 cm right, 10.0 mm tall, inverted.

Step by step solution

01

Understanding the Lens Formula

We start by recalling the lens maker's formula for a planoconvex lens:\[\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\]where \(n\) is the index of refraction, and \(R_1\) and \(R_2\) are the radii of curvature for the lens surfaces. For a planoconvex lens with one flat surface, \(R_1 = \infty\) and \(R_2 = 13.0 \ \mathrm{cm}\).
02

Calculate Focal Length

Substitute the known values into the lens maker's formula:\[\frac{1}{f} = (1.70 - 1)\left(\frac{1}{\infty} - \frac{1}{13}\right) = 0.70 \left(0 - \frac{1}{13}\right)\]This simplifies to:\[\frac{1}{f} = -\frac{0.70}{13} \Rightarrow f \approx -18.57 \ \mathrm{cm}\]
03

Calculating Image Location - Object on Left

Using the lens formula, \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(d_o = 22.5 \ \mathrm{cm}\):\[\frac{1}{-18.57} = \frac{1}{22.5} + \frac{1}{d_i}\]Solve for \(d_i\):\[\frac{1}{d_i} = \frac{1}{-18.57} - \frac{1}{22.5} \Rightarrow d_i \approx -60.00 \ \mathrm{cm}\]The negative sign indicates the image is virtual, located 60.00 cm to the right of the lens.
04

Calculating Image Size

The magnification \(M\) is given by \(M = -\frac{d_i}{d_o}\). Using \(d_i = -60.00 \ \mathrm{cm}\) and \(d_o = 22.5 \ \mathrm{cm}\):\[M = -\frac{-60.00}{22.5} \approx 2.67\]This means the image height \(h_i\) is \(3.75 \text{ mm} \times 2.67 \approx 10.0 \text{ mm}\), and it is upright.
05

Repeating with Lens Reversed

If the lens is reversed, \(R_1 = 13.0 \ \mathrm{cm}\) and \(R_2 = \infty\), repeat the calculation:\[\frac{1}{f} = (1.70 - 1)\left(\frac{1}{13} - \frac{1}{\infty}\right) = \frac{0.70}{13}\]\[f \approx 18.57 \ \mathrm{cm}\]Using the lens formula, solve for \(d_i\):\[\frac{1}{18.57} = \frac{1}{22.5} + \frac{1}{d_i} \Rightarrow d_i \approx 90.00 \ \mathrm{cm}\]The image is real, located 90.00 cm to the right of the lens, inverted, with the same magnification change leading to a height of \(10.0 \text{ mm}\).

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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 essential for understanding how lenses focus light. It provides a mathematical way to calculate the focal length of a lens based on its physical characteristics and the refractive index of the material. For a planoconvex lens, like in our exercise, the formula is given by:
  • \( \frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
In this equation:
  • \( n \) is the refractive index.
  • \( R_1 \) and \( R_2 \) are the radii of curvature for the lens surfaces.
  • A flat surface has a radius of curvature \( R = \infty \).
This format allows us to solve for the focal length \( f \), which provides insights into how the lens will converge or diverge light rays. Understanding each term's role assists in predicting the lens's behavior.
Focal Length Calculation
The focal length \( f \) of a planoconvex lens is the distance from the lens where converging light rays meet or appear to meet. Calculating the focal length is crucial for designing optical systems.Let's look at the calculation:
  • The index of refraction \( n = 1.70 \).
  • With one surface flat, \( R_1 = \infty \) and the other \( R_2 = 13.0 \text{ cm} \).
Substituting these values into the lens maker's formula:
  • \( \frac{1}{f} = (1.70 - 1)\left(0 - \frac{1}{13}\right) \).
  • Simplifies to \( f \approx -18.57 \text{ cm} \).
The negative focal length indicates a virtual focus, meaning the lens diverges light, creating a virtual illusion of light convergence.
Image Magnification
Image magnification is the measure of how much larger or smaller an image is compared to the object's actual size. It gives us crucial information about the image's appearance.For magnification \( M \), we use:
  • \( M = -\frac{d_i}{d_o} \)
Where:
  • \( d_i = -60.00 \text{ cm} \) (image distance, negative for virtual images).
  • \( d_o = 22.5 \text{ cm} \) (object distance).
The calculation gives us:
  • \( M \approx 2.67 \).
  • This means the image is upright and 2.67 times larger than the object, resulting in an image height of approximately 10.0 mm.
Lens Reversal Effects
Reversing the lens changes the focal properties significantly. In practice, this means switching the front and back surfaces of the lens, altering the radii of curvature we consider in the lens maker's formula.For a reversed planoconvex lens:
  • \( R_1 = 13.0 \text{ cm} \) and \( R_2 = \infty \).
Recalculate the focal length:
  • \( \frac{1}{f} = (1.70 - 1)\left(\frac{1}{13} - 0\right) \).
  • This changes \( f \approx 18.57 \text{ cm} \), now positive, indicating the lens is converging.
Placement of the object remains the same, but calculated image distance \( d_i \) becomes:
  • \( d_i \approx 90.00 \text{ cm} \), indicating a real, inverted image.
These changes emphasize the importance of lens orientation in optical design, as they affect both image location and orientation.

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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 2.25 times the size of the object. How far should the mirror be from the wall? What should its radius of curvature be?

Photography. A 35 -mm camera has a standard lens with focal length 50 \(\mathrm{mm}\) and can focus on objects between 45 \(\mathrm{cm}\) and infinity. (a) Is the lens for such a camera a concave or a convex lens? (b) The camera is focused by rotating the lens, which moves it on the camera body and changes its distance from the film. In what range of distances between the lens and the film plane must the lens move to focus properly over the 45 \(\mathrm{cm}\) to infinity range?

A spherical, concave shaving mirror has a radius of curvature of 32.0 \(\mathrm{cm}\) . (a) What is the magnification of a person's face when it is 12.0 \(\mathrm{cm}\) to the left of the vertex of the mirror? (b) Where is the image? Is the image real or virtual? (c) Draw a principal-ray diagram showing the formation of the image.

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?

A person with a near point of \(85 \mathrm{cm},\) but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare. (a) If the lenses of the old pair have a power of \(+2.25\) diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.0 \(\mathrm{cm}\) in front of his eye? (b) What would his near point be if his old glasses were contact lenses instead?
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