Problem 5 - University Physics with Modern Physics

Question

An object 0.600 cm tall is placed 16.5 cm to the left of the vertex of a concave spherical mirror having a radius of curvature of 22.0 cm. (a) Draw a principal-ray diagram showing the formation of the image. (b) Determine the position, size, orientation, and nature (real or virtual) of the image.

Step-by-Step Solution

Answer

The image is real, inverted, 1.2 cm tall, and located 33.0 cm to the left of the mirror.

Step 1: Understanding the Mirror Formula

For a concave mirror, we use the mirror formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length of the mirror, \( d_o \) is the object distance, and \( d_i \) is the image distance. The radius of curvature \( R = 22.0 \) cm. Since the focal length \( f \) is half of the radius of curvature for a spherical mirror, we have \( f = \frac{R}{2} = \frac{22.0}{2} = 11.0 \) cm.

Step 2: Calculating the Image Distance

Plug the known values into the mirror formula to find the image distance \( d_i \). We have \( \frac{1}{11.0} = \frac{1}{16.5} + \frac{1}{d_i} \). This simplifies to \( \frac{1}{d_i} = \frac{1}{11.0} - \frac{1}{16.5} \). Calculating this gives \( d_i = 33.0 \) cm, meaning the image forms 33.0 cm to the left of the mirror.

Step 3: Determine Image Characteristics

Use the magnification formula \( m = -\frac{d_i}{d_o} \) to find magnification. Thus, \( m = -\frac{33.0}{16.5} = -2.0 \). The negative sign indicates the image is inverted. The magnitude \( |m| = 2.0 \) tells us that the image is twice the height of the object.

Step 4: Drawing Principal Ray Diagram

In the principal ray diagram, draw three key rays: one parallel to the principal axis reflecting through the focus, another passing through the center of curvature reflecting back upon itself, and a third ray passing through the focus reflecting parallel. These rays converge at the position of the image, confirming the image distance and orientation.

Step 5: Conclusion about the Image Nature

Based on calculations and ray diagram, the image formed is real (since it is on the same side as the object and inverted), inverted, and larger than the object (image height is 1.2 cm, calculated by multiplying object height by magnification).

Key Concepts

Mirror Formula

In optics, the mirror formula connects the object distance ( \( d_o \)), the image distance ( \( d_i \) ), and the focal length ( \( f \) ) of a spherical mirror via a simple equation: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]For a concave mirror, the focal length is half of the mirror’s radius of curvature ( \( R \)). This gives us \( f = \frac{R}{2} \). If the radius of curvature is 22.0 cm, then \( f = 11.0 \) cm. This formula is essential for finding unknowns such as the image distance once we know the focal length and the object distance. Understanding this allows you to determine where an image will form and its characteristics when an object is placed in front of a concave mirror.

Image Distance Calculation

To find the image distance ( \( d_i \) ), we rearrange the mirror formula. With the object distance ( \( d_o \) ) being 16.5 cm, and the focal length ( \( f \) ) as 11.0 cm, the formula becomes:\[\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}\]Substitute the known values in:\[\frac{1}{d_i} = \frac{1}{11.0} - \frac{1}{16.5}\]By solving this, we obtain \( d_i = 33.0 \) cm. This indicates that the image forms 33.0 cm away from the mirror on the same side as the object. Calculating this distance helps visualize how close or far the image appears, and whether it is real or virtual. Since this calculation shows a positive image distance, the image is real.

Ray Diagram Construction

A ray diagram is a useful tool for visualizing and understanding image formation. For a concave mirror, several key rays are commonly drawn:

• First, a ray parallel to the principal axis, which reflects through the focal point.
• Second, a ray passing through the center of curvature, which reflects back upon itself.
• Third, a ray passing through the focal point, which reflects parallel to the principal axis.

These rays help us locate where the reflected rays meet, which is the image location. In this case, they converge at a point 33.0 cm on the object side of the mirror, confirming our calculation. Ray diagram construction not only verifies the mathematical solution but also aids in visually confirming the position and orientation of the image.

Magnification and Image Characteristics

To describe an image formed by a concave mirror, we use magnification ( \( m \) ), defined as:\[m = -\frac{d_i}{d_o}\]Substituting our values gives:\[m = -\frac{33.0}{16.5} = -2.0\]The negative sign indicates that the image is inverted. The magnitude ( \( |m|=2.0 \) ) tells us the image is twice the height of the object. Given that the original object's height is 0.600 cm, the image height is \( 1.2 \) cm. The characteristics we derive include:

• Orientation: Inverted due to the negative magnification.
• Size: The image is larger, specifically twice the size of the object.
• Nature: Real, as it forms on the same side as the object.

Understanding magnification provides insight into the size and nature of the image, completing the analysis of the image characteristics formed by the concave mirror.