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Chapter 23: Problem 95

A 5.0 -cm-tall object is placed \(50.0 \mathrm{cm}\) from a lens with focal length \(-20.0 \mathrm{cm} .\) (a) How tall is the image? (b) Is the image upright or inverted?

Short Answer

Expert verified
Question: Determine the height and orientation of the image formed by a lens when a 5 cm tall object is placed at a distance of 50 cm from the lens with a focal length of -20 cm. Answer: The height of the image is 3.33 cm, and the image is upright.

Step by step solution

01

1. Identify the given parameters

In the problem, we are given the following information: - Object height (h_o) = 5 cm - Object distance (d_o) = 50 cm - Focal length (f) = -20 cm
02

2. Apply the Thin Lens Equation

The Thin Lens equation is given by: \(\mathrm{\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}}\) We need to determine the image distance (d_i) from the given information. Just plug in the values and solve for \(d_i\): \(\mathrm{\frac{1}{-20} = \frac{1}{50} + \frac{1}{d_i}}\) Now, find \(d_i\) by solving the equation.
03

3. Calculate Image Distance (d_i)

To find d_i, rewrite the equation as follows: \(\mathrm{\frac{1}{d_i} = \frac{1}{-20} - \frac{1}{50}}\) Next, find a common denominator and combine the fractions: \(\frac{1}{d_i} = \frac{-3}{100}\) To find \(d_i\), take the reciprocal of both sides: \(d_i = -\frac{100}{3} 鈮� -33.33 \thinspace cm\) So, the image is formed at a distance of approximately -33.33 cm from the lens.
04

4. Calculate Image Height (h_i) using Magnification Equation

The magnification (M) equation is given by: \(M = \frac{h_i}{h_o} = \frac{-d_i}{d_o}\) We can derive the image height (h_i) from the equation as: \(h_i = M \times h_o\) Now, calculating the magnification (M): \(M = \frac{-d_i}{d_o} = \frac{-(-33.33)}{50} = \frac{33.33}{50} = 0.67\) Finally, calculate the image height (h_i): \(h_i = M \times h_o = 0.67 \times 5 = 3.33 \thinspace cm\) Thus, the height of the image is 3.33 cm.
05

5. Determine Image Orientation

Now that we've found the height of the image, we can determine its orientation. Since the magnification is positive (M = 0.67), the image is upright, meaning that it is oriented in the same direction as the object. In conclusion: (a) The image is 3.33 cm tall. (b) The image is upright.

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Most popular questions from this chapter

A penny is at the bottom of a bowl full of water. When you look at the water surface from the side, with your eyes at the water level, the penny appears to be just barely under the surface and a horizontal distance of $3.0 \mathrm{cm}\( from the edge of the bowl. If the penny is actually \)8.0 \mathrm{cm}$ below the water surface, what is the horizontal distance between the penny and the edge of the bowl? [Hint: The rays you see pass from water to air with refraction angles close to \(\left.90^{\circ} .\right]\)
An object of height \(5.00 \mathrm{cm}\) is placed \(20.0 \mathrm{cm}\) from a converging lens of focal length \(15.0 \mathrm{cm} .\) Draw a ray diagram to find the height and position of the image.
An object is placed \(10.0 \mathrm{cm}\) in front of a lens. An upright, virtual image is formed \(30.0 \mathrm{cm}\) away from the lens. What is the focal length of the lens? Is the lens converging or diverging?
An object of height \(3.00 \mathrm{cm}\) is placed \(12.0 \mathrm{cm}\) from a diverging lens of focal length \(-12.0 \mathrm{cm} .\) Draw a ray diagram to find the height and position of the image.
A diverging lens has a focal length of \(-8.00 \mathrm{cm} .\) (a) What are the image distances for objects placed at these distances from the lens: $5.00 \mathrm{cm}, 8.00 \mathrm{cm}, 14.0 \mathrm{cm}, 16.0 \mathrm{cm}$ \(20.0 \mathrm{cm} ?\) In each case, describe the image as real or virtual, upright or inverted, and enlarged or diminished in size. (b) If the object is \(4.00 \mathrm{cm}\) high, what is the height of the image for the object distances of \(5.00 \mathrm{cm}\) and \(20.0 \mathrm{cm} ?\)
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