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A lens forms an image of an object. The object is 16.0 cm from the lens. The image is 12.0 cm from the lens on the same side as the object. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.50 mm tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.

Short Answer

Expert verified
(a) Focal length \(f = -48\, \text{cm}\), diverging lens. (b) Image height is \(-6.375\, \text{mm}\), image is inverted.

Step by step solution

01

Identify the Given Information

We are given that the object distance \(d_o = 16.0\, \text{cm}\) and the image distance \(d_i = -12.0\, \text{cm}\). The negative sign indicates the image is on the same side as the object, implying the image is virtual.
02

Use Lens Formula for Focal Length

The lens formula is \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). Substitute the known values: \(\frac{1}{f} = \frac{1}{16.0} + \frac{1}{-12.0}\). Simplifying this gives \(\frac{1}{f} = \frac{1}{16.0} - \frac{1}{12.0}\). Calculate \(\frac{1}{f}\) to determine \(f\).
03

Calculate Focal Length

Convert the equation to find \(f\): \(\frac{1}{f} = \frac{1}{16.0} - \frac{1}{12.0} = \frac{3}{48} - \frac{4}{48} = -\frac{1}{48}\). Thus, \(f = -48.0 \text{ cm}\). Since the focal length is negative, the lens is a diverging lens.
04

Determine Image Height Using Magnification

The magnification \(m\) is given by \(m = \frac{h_i}{h_o} = \frac{d_i}{d_o}\), where \(h_o = 8.50\, \text{mm}\) and \(d_i = -12.0\, \text{cm}\), \(d_o = 16.0\, \text{cm}\). This means \(m = \frac{-12.0}{16.0}\). Calculate \(m\) to find \(h_i\).
05

Calculate Image Height and Determine Its Orientation

Using the magnification, \(m = -\frac{3}{4}\), and \(h_o = 8.50\, \text{mm}\), we determine \(h_i = m \times h_o = -\frac{3}{4} \times 8.50 = -6.375\, \text{mm}\). The negative sign indicates the image is inverted.

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Key Concepts

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

Focal Length
The focal length of a lens indicates how strongly it converges or diverges light. It is the distance between the lens and its focus, where parallel rays of light converge (or appear to diverge from). The lens formula can be expressed as \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance.
In this exercise, the negative image distance tells us the image is virtual and on the same side as the object.
After substituting the given values, we found \( f = -48.0 \, \text{cm} \), indicating a diverging lens. A negative focal length characterizes lenses that spread out light rays, instead of bringing them together.
Magnification
Magnification tells us how much larger or smaller the image is compared to the object. It is calculated using the formula \( m = \frac{h_i}{h_o} = \frac{d_i}{d_o} \), where \( h_i \) is the image height and \( h_o \) is the object height. For instance, in our example, the object was 8.50 mm tall.
The magnification was found to be \( m = -\frac{3}{4} \), meaning the image's height is three quarters that of the object, and its inverted position results in a negative value.
The greater the absolute value of magnification, the larger the image appears relative to the object.
  • If \( m > 1 \), the image is larger than the object.
  • If \( m = 1 \), the image and object have the same size.
  • If \( m < 1 \), the image is smaller than the object.
Diverging Lens
A diverging lens, often called a concave lens, has at least one surface that curves inward.
This type of lens causes incoming parallel light rays to spread out as if they were emanating from a common point (the focal point) on the opposite side of the lens.
Diverging lenses are known for forming virtual images, as the rays appear to spread out from a point on the same side as the object.
  • They have a negative focal length.
  • Are used in applications such as correcting short-sightedness.
Understanding the behavior of diverging lenses helps in comprehending how optical instruments, such as cameras and glasses, manipulate light.
Virtual Image
A virtual image is formed when light rays diverge, causing the image to appear on the same side as the object. In this case, the image cannot be captured on a screen because the light does not actually converge to create the image.
Virtual images, like the one in our problem, are erect if viewed through the lens but appear inverted due to calculated magnification. They typically have the following characteristics:
  • Cannot be displayed on a screen.
  • Appear where light rays only seem to converge.
  • Commonly produced by diverging lenses and flat mirrors.
Understanding these images plays a crucial role in designing optical devices, training in optics lab practices, and solving optical puzzles.

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Most popular questions from this chapter

The eyepiece of a refracting telescope (see Fig. 34.53) has a focal length of 9.00 cm. The distance between objective and eyepiece is 1.20 m, and the final image is at infinity. What is the angular magnification of the telescope?

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.

An object is 18.0 cm from the center of a spherical silvered-glass Christmas tree ornament 6.00 cm in diameter. What are the position and magnification of its image?

In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the bending occurs at the cornea, whose vertex is 2.60 cm from the retina. What should be the radius of curvature of the cornea such that the image of an object 40.0 cm from the cornea's vertex is focused on the retina?

A 1.20-cm-tall object is 50.0 cm to the left of a converging lens of focal length 40.0 cm. A second converging lens, this one having a focal length of 60.0 cm, is located 300.0 cm to the right of the first lens along the same optic axis. (a) Find the location and height of the image (call it \(I_1\)) formed by the lens with a focal length of 40.0 cm. (b) \(I_1\) is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.

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