Question

An object is \(6.0 \mathrm{~cm}\) from a converging thin lens along the axis of the lens. If the lens has a focal length of \(9.0 \mathrm{~cm},\) determine the image magnification.

Short answer

Question: An object is placed 6.0 cm from a converging lens with a focal length of 9.0 cm. Calculate the magnification of the image. Answer: The magnification of the image is -3, which means the image is 3 times larger than the object and inverted.

Step-by-step solution

Understand and state the given values

An object is placed 6.0 cm from a converging lens, with a focal length of 9.0 cm. We'll use these values to find the image magnification. First, let's convert the values into meters. Object distance, \(s_o = 6.0\,\text{cm} = 0.06\,\text{m}\) Focal length, \(f = 9.0\,\text{cm} = 0.09\,\text{m}\)

Lens formula

The lens formula is given by the equation: \(\frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i}\) Where \(f\) is the focal length, \(s_o\) is the object distance, and \(s_i\) is the image distance. We'll use this formula to find the image distance, \(s_i\).

Calculate the image distance

Rearrange the lens formula to solve for \(s_i\): \(s_i = \frac{1}{\frac{1}{f} - \frac{1}{s_o}}\) Plug in the known values of \(f\) and \(s_o\) to find \(s_i\): \(s_i = \frac{1}{\frac{1}{0.09} - \frac{1}{0.06}} \Rightarrow s_i = 0.18\,\text{m}\) The image is formed at a distance of 0.18 meters from the lens.

Magnification formula

Now that we have the image distance, we can find the image magnification using the magnification formula: \(m = -\frac{s_i}{s_o}\)

Calculate the magnification

Plug in the values of \(s_i\) and \(s_o\) into the magnification formula to find the magnification: \(m = -\frac{0.18}{0.06} \Rightarrow m = -3\) The magnification of the image is -3, which means the image is 3 times larger than the object and inverted (due to the negative sign).

Key concepts

Understanding Converging Lenses

A converging lens is crafted to bend incoming parallel light rays toward a centerpiece known as the focal point. These lenses are thicker in the middle compared to their edges, giving them their classic 'biconvex' shape. An example of a converging lens is a magnifying glass, which uses its focal point to enlarge objects when they are placed between the focal point and the lens.

When an object is placed at a distance from a converging lens, the light from that object passes through the lens and either converges or appears to diverge from a specific point, forming an image. The type of image, whether it is real or virtual, largely depends on the object's distance relative to the lens's focal length. When the object is beyond the focal length, as is the case in our example problem, a real and inverted image is formed on the other side of the lens.

Applying the Lens Formula

The lens formula is a powerful tool in optics that relates the focal length of a lens (f), the distance from the lens to the object (s_o), and the distance from the lens to the image (s_i). It is generally expressed as:
\[ \frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i} \]
By rearranging the formula, you can solve for any one of the variables when the other two are known. The lens formula is applicable for any thin lens, whether it’s converging or diverging; however, you need to keep track of the sign convention. Generally, for converging lenses, the focal length is positive, and when calculating for the image distance, if you get a positive result, it indicates a real image is formed on the opposite side of the object.

Deciphering the Magnification Formula

Magnification, denoted by the letter 'm', indicates how much larger or smaller the image is compared to the object. It can be positive or negative, which also tells us the orientation of the image. We use the magnification formula:
\[ m = -\frac{s_i}{s_o} \]
This formula is derived from the proportions of similar triangles formed by the object and the image in relation to the lens. If the magnification is negative, as in our exercise, this indicates that the image is inverted relative to the object. When calculating the magnification for our example, where the image distance (s_i) is thrice the object distance (s_o), we find that the image is three times larger and inverted - a key point to remember in visualizing and understanding the outcome of the lens equation in practical applications.