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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: \(d_i = -\frac{100}{3} ≈ -33.33 \thinspace cm\) 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 glass lens has a scratch-resistant plastic coating on it. The speed of light in the glass is \(0.67 c\) and the speed of light in the coating is \(0.80 c .\) A ray of light in the coating is incident on the plastic-glass boundary at an angle of \(12.0^{\circ}\) with respect to the normal. At what angle with respect to the normal is the ray transmitted into the glass?
A diamond in air is illuminated with white light. On one particular facet, the angle of incidence is \(26.00^{\circ} .\) Inside the diamond, red light \((\lambda=660.0 \mathrm{nm}\) in vacuum) is refracted at \(10.48^{\circ}\) with respect to the normal; blue light $(\lambda=470.0 \mathrm{nm} \text { in vacuum })\( is refracted at \)10.33^{\circ} .$ (a) What are the indices of refraction for red and blue light in diamond? (b) What is the ratio of the speed of red light to the speed of blue light in diamond? (c) How would a diamond look if there were no dispersion?
What is the index of refraction of the core of an optical fiber if the cladding has \(n=1.20\) and the critical angle at the core-cladding boundary is \(45.0^{\circ} ?\)
(a) Calculate the critical angle for a diamond surrounded by air. (b) Calculate the critical angle for a diamond under water. (c) Explain why a diamond sparkles less under water than in air.
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.
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