Question

A \(5.00-\mathrm{cm}\) object is placed \(30.0 \mathrm{~cm}\) away from a convex mirror with a focal length of \(-10.0 \mathrm{~cm}\). Determine the size, orientation, and position of the image.

Short answer

Answer: The image formed by the convex mirror has a size of 2.50 cm, is virtual and upright, and is positioned on the same side as the object, 15 cm away from the mirror.

Step-by-step solution

Identify the given information

The given information is: - Object distance (d_o): 30.0 cm - Object height (h_o): 5.00 cm - Focal length (f): -10.0 cm

Write the mirror equation and magnification equation

The mirror equation is: \(\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i} \) The magnification (m) equation is: \(m=\frac{-d_i}{d_o}=\frac{h_i}{h_o}\)

Solve for the image distance (d_i)

We'll start by solving for the image distance using the mirror equation: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) Plugging in the values: \(\frac{1}{-10} = \frac{1}{30} + \frac{1}{d_i} \) Now, isolate \(d_i\): \(\frac{1}{d_i} = \frac{1}{-10} - \frac{1}{30} \) To find the common denominator: \(\frac{1}{d_i} = \frac{-3+1}{30} = \frac{-2}{30} \) Take the reciprocal of both sides: \( d_i = \frac{30}{-2} = -15 \mathrm{~cm} \) The image distance is -15 cm.

Determine the image orientation

Since the image distance (d_i) is negative, the image is virtual and formed on the same side as the object.

Solve for image height (h_i) using magnification equation

We can use the magnification equation to find the image height: \(m = \frac{-d_i}{d_o} \) Use the image distance we found previously: \(m = \frac{-(-15)}{30} = \frac{1}{2} \) Now use the second part of the magnification equation to find the image height: \(m = \frac{h_i}{h_o} \) Plugging in the values: \(\frac{1}{2} = \frac{h_i}{5.00} \) Now, solve for \(h_i\): \(h_i = \frac{1}{2} \times 5.00 = 2.50 \mathrm{~cm} \) The image height is 2.50 cm.

Write down the final answers

The image formed by the convex mirror has the following properties: - Size: 2.50 cm (smaller than the object) - Orientation: Virtual and upright (since d_i is negative) - Position: -15 cm (on the same side as the object and 15 cm away from the mirror)

Key concepts

Mirror Equation

The mirror equation is a fundamental tool in understanding how images are formed by mirrors. In a convex mirror, the focal length (f) is negative because it diverges light rays. When we apply the mirror equation, \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), it’s important to remember that each term represents the inverse of a distance: the focal length (f), object distance (\(d_o\)), and image distance (\(d_i\)).

In the given problem, by substituting the provided values into the mirror equation and solving for \(d_i\), we learned that the image distance is negative. This negative sign is indicative of a virtual image, as it implies that the image is formed on the same side as the object -- a key characteristic of convex mirrors. This property means that the image is not real; it cannot be projected onto a screen, as opposed to a real image formed by concave mirrors where light actually converges.

Magnification Equation

The magnification equation, \( m = \frac{-d_i}{d_o} = \frac{h_i}{h_o} \), reveals much about the image created by a mirror. It consists of two ratios: the first dealing with distances (image to object distance), and the second comparing heights (image to object height).

A magnification (m) value greater than one indicates an enlargement, while a value less than one, as found in our problem, means the image is smaller than the object. Furthermore, the negative sign in the magnification equation for mirrors signifies the orientation of the image. In convex mirrors, the outcome will always be positive, reflecting the upright nature of the image. This was demonstrated when we solved for the image height (\(h_i\)), obtaining a value that was positive and smaller than the object height (\(h_o\)), confirming the equation's indication of a smaller, upright virtual image.

Virtual Image

A virtual image, intriguingly, is a concept rather than a tangible reality. It is where light rays appear to diverge from, as opposed to a real image where they truly converge. Due to the diverging property of convex mirrors, they only form virtual images. These images are always upright, scaled down or up from the object size, and are formed on the same side as the object itself. This is why the image distance in our exercise is negative, signifying its virtual nature. Unlike real images, virtual ones cannot be captured on a screen, yet they are visible to us because our eyes trace the light rays back to their apparent origin, hence 'seeing' the virtual image.

Object-Image Distance Relationship

The relationship between object and image distance in the context of convex mirrors is explained by understanding that as the object moves closer to the mirror, the image distance decreases. Conversely, as the object moves further away, the image appears closer to the mirror. This relationship is bound by the mirror's focal length, which remains constant for a particular mirror. The result is a direct and mathematically predictable relation, essential for precisely calculating where an image will appear relative to the mirror. In our example, we found a virtual image located 15 cm from the mirror, which, compared to a real object 30 cm away, gives a practical sense of how convex mirrors manipulate perceived distance.