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Chapter 24: Problem 10

You plan to project an inverted image \(30.0 \mathrm{cm}\) to the right of an object. You have a diverging lens with focal length \(-4.00 \mathrm{cm}\) located \(6.00 \mathrm{cm}\) to the right of the object. Once you put a second lens at \(18.0 \mathrm{cm}\) to the right of the object, you obtain an image in the proper location. (a) What is the focal length of the second lens? (b) Is this lens converging or diverging? (c) What is the total magnification? (d) If the object is \(12.0 \mathrm{cm}\) high. what is the image height?

Short Answer

Expert verified
Question: In a two-lens system, the first lens is a diverging lens with a focal length of -4.00 cm. An object is placed 6.00 cm from this lens, and the lenses are separated by 12 cm. If the final image is inverted and 30 cm to the right of the object, find (a) the focal length and type of the second lens, (b) the total magnification of the system, and (c) the image height if the object height is 12.0 cm. Answer: (a) The focal length of the second lens is 8 cm, and it is a converging lens. (b) The total magnification of the system is -1. (c) The image height is -12 cm, which means the image is inverted and the same size as the object.

Step by step solution

01

Use the thin lens formula for the first lens

The Thin Lens Formula is given by: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] where: - \(f\) is the focal length of the lens, - \(d_o\) is the object distance (distance between object and lens), - \(d_i\) is the image distance (distance between image and lens). For the first lens, we know the object distance (\(d_o = 6.00 cm\)) and the focal length (\(f = -4.00 cm\)). Let's calculate the image distance: \[ \frac{1}{-4} = \frac{1}{6} + \frac{1}{d_{i1}} \] \[ d_{i1} = -12 cm \] So, the image distance for the first lens is \(-12 cm\).
02

Calculate the object distance for the second lens

The object distance for the second lens is the distance between the first lens and the second lens plus the virtual image distance for the first lens. The distance between the two lenses is given as 12 cm, and we know the virtual image distance for the first lens is -12 cm: \[d_{o2} = 12 - (-12) = 24 cm\]
03

Calculate the image distance for the second lens

The total distance of the inverted image is 30 cm to the right of the object. We know that the second lens is 18 cm to the right of the object. Therefore, the image distance for the second lens is: \[ d_{i2} = 30 - 18 = 12 cm \]
04

Use the thin lens formula for the second lens to find its focal length

We now know the object distance and image distance for the second lens, so we can apply the Thin Lens Formula to find its focal length: \[ \frac{1}{f_2} = \frac{1}{d_{o2}} + \frac{1}{d_{i2}} \] \[ \frac{1}{f_2} = \frac{1}{24} + \frac{1}{12} \] \[ f_2 = 8 cm \] (a) The focal length of the second lens is \(8 cm\).
05

Identify the type of the second lens

Since the focal length of the second lens is positive, it is a converging lens. (b) The second lens is converging.
06

Calculate the total magnification of the system

To find the total magnification, we calculate the magnification for each lens individually, and then multiply them: Magnification equation: \[ M_i = -\frac{d_{i}}{d_{o}} \] Lens 1: \[ M_1 = -\frac{-12}{6} = 2\] Lens 2: \[ M_2 = -\frac{12}{24} = -0.5\] Now, multiply the magnifications to get the total magnification: \[ M_{total} = M_1 \times M_2 = 2 \times (-0.5) = -1 \] (c) The total magnification of the system is -1.
07

Calculate the image height

Now that we have calculated the total magnification of the two-lens system, we can use the magnification formula to determine the image height: \[ M_{total} = \frac{h_{i}}{h_{o}} \] The object height is given as \(12.0 cm\). Plug in the given values and solve for the image height: \[-1 = \frac{h_i}{12} \] \[ h_i = -12 cm \] (d) The image height is -12 cm, which means the image is inverted and the same size as the object.

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Most popular questions from this chapter

Two converging lenses are placed \(88.0 \mathrm{cm}\) apart. An object is placed \(1.100 \mathrm{m}\) to the left of the first lens. which has a focal length of \(25.0 \mathrm{cm} .\) The final image is located \(15.0 \mathrm{cm}\) to the right of the second lens. (a) What is the focal length of the second lens? (b) What is the total magnification?
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