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Chapter 33: Problem 91

An instructor wants to use a lens to project a real image of a light bulb onto a screen \(1.71 \mathrm{~m}\) from the bulb. In order to get the image to be twice as large as the bulb, what focal length lens will be needed?

Short Answer

Expert verified
Answer: The focal length of the lens needed is approximately 1.14 meters.

Step by step solution

01

Define Variables

Let's denote the focal length as \(f\), the object distance as \(d_o\), the image distance as \(d_i\), and the magnification as \(m\). In this problem, we have: \(d_o = 1.71m\) \(m = 2\)
02

Using the Lens Maker's Equation

The lens maker's equation is given by: \(\frac{1}{f} = \frac{1}{d_i} + \frac{1}{d_o}\), where \(f\) is the focal length, \(d_i\) is the image distance, and \(d_o\) is the object distance. We want to find the value of \(f\).
03

Using the Magnification Equation

The magnification equation is given by: \(m = \frac{d_i}{d_o}\). Since we want the magnification to be \(2\), and the object distance is \(1.71m\), we can find the image distance \(d_i\): \(m = 2 = \frac{d_i}{1.71}\) \(d_i = 2 \times 1.71 = 3.42m\)
04

Substituting Values into the Lens Maker's Equation

Now that we have the values for \(d_o\) and \(d_i\), we can plug them into the lens maker's equation to find the focal length \(f\): \(\frac{1}{f} = \frac{1}{3.42} + \frac{1}{1.71}\) \(\frac{1}{f} = 0.2923 + 0.5848\) \(\frac{1}{f} = 0.8771\)
05

Solving for the Focal Length

Now, we'll solve for the focal length by taking the reciprocal of the resulting value: \(f = \frac{1}{0.8771}\) \(f \approx 1.14m\) The focal length of the lens needed is approximately \(1.14m\).

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Most popular questions from this chapter

What type of lens is a magnifying glass? a) converging d) cylindrical b) diverging e) plain c) spherical

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