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Chapter 23: Problem 63

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.

Short Answer

Expert verified
Question: An object with a height of 3.00 cm is placed 12.0 cm from a diverging lens with a focal length of -12.0 cm. Find the height and position of the image formed by the lens. Answer: The image is a virtual image located 4.00 cm to the left of the lens and has a height of 1.00 cm.

Step by step solution

01

Identify the object's properties

The object has a height of \(3.00 \mathrm{cm}\) and is placed \(12.0 \mathrm{cm}\) from the diverging lens. The lens has a focal length of \(-12.0 \mathrm{cm}\).
02

Find the focal point of the lens

A diverging lens has a negative focal length, so the focal point is on the same side as the object. In this case, the focal point is \(12 \mathrm{cm}\) to the left of the lens.
03

Draw the ray diagram

Follow these rules to draw the ray diagram: 1. Draw a straight line from the object's top point, parallel to the principal axis, until it reaches the lens. After passing through the lens, this ray will diverge as if it's coming from the focal point. 2. Draw another straight line from the top point of the object through the center of the lens. This ray will continue undeviated. 3. The image is formed where these two rays appear to intersect.
04

Use the lens formula to find the image distance

The lens formula is: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) We know the focal length \(f = -12.0 \mathrm{cm}\) and the object distance \(d_o = 12.0 \mathrm{cm}\). So we can solve for the image distance \(d_i\): \(\frac{1}{-12} = \frac{1}{12} + \frac{1}{d_i} \Rightarrow d_i = -4 \mathrm{cm}\) The negative sign means that the image is virtual and located on the same side of the lens as the object. The image is \(4 \mathrm{cm}\) to the left of the lens.
05

Calculate the image height using the magnification formula

The magnification formula is: \(m = \frac{h_i}{h_o} = \frac{d_i}{d_o}\) We know the object's height \(h_o = 3.00 \mathrm{cm}\), the object distance \(d_o = 12.0 \mathrm{cm}\), and the image distance \(d_i = -4 \mathrm{cm}\). We can now find the image height \(h_i\): \(-\frac{h_i}{3} = \frac{-4}{12} \Rightarrow h_i = 1.00 \mathrm{cm}\) The image height is \(1.00 \mathrm{cm}\). Since the magnification is negative, the image is inverted. So, the image is a virtual image located \(4.00 \mathrm{cm}\) to the left of the lens and has a height of \(1.00 \mathrm{cm}\).

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