Problem 10 - University Physics with Modern Physics

Question

You hold a spherical salad bowl 90 \(\mathrm{cm}\) in front of your face with the bottom of the bowl facing you. The salad bowl is made of polished metal with a \(35-\) cm radius of curvature. (a) Where is the image of your 2.0 -cm- tall nose located? (b) What are the image's size, orientation, and nature (real or virtual)?

Step-by-Step Solution

Answer

The image is located 22.64 cm behind the mirror, 0.503 cm tall, upright, and virtual.

Step 1: Understand the Mirror Equation

The problem involves a spherical mirror. The mirror equation relates the object distance \(d_o\), the image distance \(d_i\), and the focal length \(f\) of a mirror: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). For a spherical mirror, the focal length \(f\) is half the radius of curvature \(R\), so \(f = \frac{R}{2}\).

Step 2: Determine the Focal Length

Given the radius of curvature \(R = 35\, \text{cm}\), calculate the focal length: \(f = \frac{35}{2} = 17.5\, \text{cm}\).

Step 3: Use the Mirror Equation to Find Image Distance

The object distance \(d_o\) is the distance you hold the bowl from your face, which is given as \(90\, \text{cm}\). Apply the mirror equation: \(\frac{1}{17.5} = \frac{1}{90} + \frac{1}{d_i}\). Solve for \(d_i\).

Step 4: Solve for Image Distance

Rearrange the equation to find \(d_i\): \(\frac{1}{d_i} = \frac{1}{17.5} - \frac{1}{90}\). Compute \(d_i\) using these values to find that \(d_i \approx -22.64\, \text{cm}\). The negative sign indicates that the image is virtual.

Step 5: Determine Image Properties Using the Magnification Equation

The magnification \(m\) of a mirror is given by \(m = -\frac{d_i}{d_o}\). The value of \(m\) can be used to determine the size of the image. Also, magnification indicates the orientation: if \(m > 0\), the image is upright; if \(m < 0\), it is inverted.

Step 6: Calculate Magnification and Image Size

Substitute \(d_i = -22.64\, \text{cm}\) and \(d_o = 90\, \text{cm}\) into the magnification equation: \(m = -\frac{-22.64}{90} \approx 0.2516\). The height of the image is \(\text{image height} = m \times 2.0\, \text{cm} \approx 0.5032\, \text{cm}\). The image is upright and virtual because \(m > 0\).

Key Concepts

Mirror Equation and How It Works

The mirror equation is a fundamental tool for solving problems with spherical mirrors. It connects three important quantities in mirror physics: the object distance \(d_o\), the image distance \(d_i\), and the focal length \(f\) of a mirror. The mirror equation is expressed as:

• \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\)

This equation helps us find where the image will form relative to the mirror.
In our case of the spherical salad bowl (acting as a concave mirror), the radius of curvature \(R\) is given as 35 cm. The focal length \(f\) of a spherical mirror is half of its radius of curvature:

• \(f = \frac{R}{2}\)
• \(f = \frac{35}{2} = 17.5\, \text{cm}\)

With this information, we can use the mirror equation to find the image distance \(d_i\) from known values, such as the object distance \(d_o\) which, in this exercise, is 90 cm.
Solving the equation \(\frac{1}{17.5} = \frac{1}{90} + \frac{1}{d_i}\) gives us \(d_i \approx -22.64\, \text{cm}\), which means the image forms on the same side as the object and is virtual.

Understanding Virtual Images in Spherical Mirrors

In optics, a virtual image is formed when the outgoing rays from a point on an object diverge. Instead of converging to a point, they appear to come from a point behind the mirror. This explains why the image cannot be projected onto a screen. Virtual images are always upright with respect to the object.
Let's relate this to our exercise. The computed image distance, \(d_i = -22.64 \text{ cm}\), being negative, confirms that the image is virtual. It's formed behind the actual reflecting surface of the bowl.
Unlike real images, which can be captured on a screen, virtual images exist because our eyes or cameras perceive the direction from which light seems to come. They appear larger, smaller, or the same size as the object, depending on the object distance and focal length.

• Virtual images often appear larger if the object is closer to the focal point.
• If the object is beyond the focal point, a real and inverted image can form.

Since the dealt image here is virtual, it appears upright and located on the side opposite of light reflection.

Magnification Equation and Image Size

The magnification equation is crucial to determining how large or small an image appears compared to the actual object. It is represented by the formula:

• \(m = -\frac{d_i}{d_o}\)

Where \(m\) is the magnification, \(d_i\) is the image distance, and \(d_o\) is the object distance.
Magnification tells us more than just size. It indicates whether the image is inverted or upright:

• If \(m > 0\), the image is upright.
• If \(m < 0\), it is inverted.

In this exercise, substituting the values \(d_i = -22.64\, \text{cm}\) and \(d_o = 90\, \text{cm}\) into the magnification equation gives us:

• \(m = -\frac{-22.64}{90} \approx 0.2516\)

This positive magnification confirms the image is upright. To find the image size, multiply the magnification by the object size (where the original nose height is 2.0 cm):

• \(\text{Image height} = 0.2516 \times 2.0 = 0.5032\, \text{cm}\)

So, the virtual image of the nose will appear at 0.5032 cm tall, significantly smaller than the actual object, further supporting the formation of a virtual, upright image.