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Chapter 2: Problem 9

Under what circumstances will an image be located at the focal point of a spherical lens or mirror?

Short Answer

Expert verified
An image cannot be located at the focal point of a spherical lens. For a spherical mirror, an image will be located at the focal point when the object distance is half of the focal length (d_o = f/2).

Step by step solution

01

The lens formula relates the object distance (d_o), image distance (d_i), and focal length (f) of a lens: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) Similarly, for a mirror, the formula is: \( \frac{1}{f} = \frac{1}{d_o} - \frac{1}{d_i} \) These formulas can be used to find out the relationship between the object, image, and focal points for a lens or a mirror. #Step 2: Determine when the image is located at the focal point#

The image is located at the focal point when the image distance (d_i) is equal to the focal length (f). Thus, we need to find out the object distance (d_o) under these circumstances for both lenses and mirrors. For a lens: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{f} \) For a mirror: \( \frac{1}{f} = \frac{1}{d_o} - \frac{1}{f} \) #Step 3: Solve for object distance (d_o)#
02

Now, we can solve these equations for object distance (d_o) in the case of the lens and the mirror. For a lens: \( \frac{1}{d_o} = \frac{1}{f} - \frac{1}{f} = 0 \) Since the reciprocal of 0 is undefined, there is no object distance for a lens where the image is located at the focal point. For a mirror: \( \frac{1}{d_o} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f} \) \( d_o = \frac{f}{2} \) #Step 4: Interpret the results#

Based on our calculations, an image cannot be located at the focal point of a spherical lens. However, for a spherical mirror, an image is located at the focal point when the object distance is half of the focal length (d_o = f/2).

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Most popular questions from this chapter

An object of height \(3.0 \mathrm{cm}\) is placed \(5.0 \mathrm{cm}\) in front of a converging lens of focal length \(20 \mathrm{cm}\) and observed from the other side. Where and how large is the image?

Where should a 3 cm tall object be placed in front of a concave mirror of radius \(20 \mathrm{cm}\) so that its image is real and \(2 \mathrm{cm}\) tall?

The power for normal close vision is 54.0 D. In a vision-correction procedure, the power of a patient's eye is increased by 3.00 D. Assuming that this produces normal close vision, what was the patient's near point before the procedure?

Referring to the electric room heater considered in problem \(5,\) calculate the intensity of IR radiation in \(\mathrm{W} / \mathrm{m}^{2}\) projected by the concave mirror on a person \(3.00 \mathrm{m}\) away. Assume that the heating element radiates \(1500 \mathrm{W}\) and has an area of \(100 \mathrm{cm}^{2}\), and that half of the radiated power is reflected and focused by the mirror.

Show that, for a flat mirror, \(h_{i}=h_{\mathrm{o}},\) given that the image is the same distance behind the mirror as the distance of the object from the mirror.
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