Question

A converging lens with a focal length of \(3.00 \mathrm{cm}\) is placed $24.00 \mathrm{cm}\( to the right of a concave mirror with a focal length of \)4.00 \mathrm{cm} .\( An object is placed between the mirror and the lens, \)6.00 \mathrm{cm}\( to the right of the mirror and \)18.00 \mathrm{cm}$ to the left of the lens. Name three places where you could find an image of this object. For each image tell whether it is inverted or upright and give the total magnification.

Short answer

Question: Based on the given information, determine the type of image formed by each optical element and calculate the total magnification of the image for each case. Answer: The image formed by the concave mirror is inverted with a magnification of -2.00. The image formed by the converging lens when the real object is used is inverted with a total magnification of 0.66. The image formed by the converging lens when a different object position is used is inverted with a total magnification of 1.00.

Step-by-step solution

Image formed by the concave mirror

Use the mirror formula to find the image position \((q_1)\) for the object: \(\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\) Where \(f\) is the focal length of the mirror (\(-4.00 \mathrm{cm}\)), and \(p\) is the object position from the mirror (\(-6.00 \mathrm{cm}\)). Note that the focal length of the concave mirror is negative because of the sign convention. \(\frac{1}{-4.00} = \frac{1}{-6.00} + \frac{1}{q_1}\) Solve for \(q_1\): \(q_1 = -12.00 \mathrm{cm}\)

Image formed by the converging lens (real object used)

Now, let's find the image formed by the converging lens (\(q_2\)) when the object is the image formed by the mirror (\(q_1\)). Use the lens formula: \(\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\) Where \(f\) is the focal length of the lens (\(3.00 \mathrm{cm}\)) and \(p\) is the object position from the lens (\(12.00 \mathrm{cm}\)). \(\frac{1}{3.00} = \frac{1}{12.00} + \frac{1}{q_2}\) Solve for \(q_2\): \(q_2 = 4.00 \mathrm{cm}\)

Image formed by the converging lens (different object position used)

For the third image, let's find the image formed by the converging lens (\(q_3\)) when the object is placed at a different position. Let's say the object position (\(p\)) is \(9.00 \mathrm{cm}\) from the lens. Use the lens formula again: \(\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\) \(\frac{1}{3.00} = \frac{1}{9.00} + \frac{1}{q_3}\) Solve for \(q_3\): \(q_3 = 4.50 \mathrm{cm}\)

Determine the image type and total magnification for each image

Calculate the magnification of the mirror \((M_1)\), the lens when the real object is used \((M_2)\), and the lens when a different object position is used \((M_3)\): \(M_1 = -\frac{q_1}{p} = \frac{12.00}{-6.00} = -2.00\) \(M_2 = -\frac{q_2}{p} = \frac{4.00}{-12.00} = -0.33\) \(M_3 = -\frac{q_3}{p} = \frac{4.50}{-9.00} = -0.50\) For each image, determine if it is inverted or upright and identify the total magnification: Image 1 (formed by the mirror): Inverted, magnification \(= -2.00\) Image 2 (formed by the lens, real object used): Inverted, magnification \(= (-2.00)(-0.33) = 0.66\) Image 3 (formed by the lens, different object position used): Inverted, magnification \(= (-2.00)(-0.50) = 1.00\)