Problem 11 - College Physics

Question

An object is \(100 \mathrm{~cm}\) in front of a concave mirror that has a radius of \(80 \mathrm{~cm} .\) (a) Use a ray diagram to determine whether the image is (1) real or virtual, (2) upright or inverted, and (3) magnified or reduced. (b) Calculate the image distance and lateral magnification.

Step-by-Step Solution

Answer

(a) The image is real, inverted, and reduced. (b) Image distance: 66.67 cm; magnification: -0.667.

Step 1: Determine the Mirror's Focal Length

The focal length \( f \) of a concave mirror can be calculated using the relation between the radius of curvature \( R \) and the focal length \( f = \frac{R}{2} \). Given \( R = 80 \mathrm{~cm} \), we find \( f = \frac{80 \mathrm{~cm}}{2} = 40 \mathrm{~cm} \).

Step 2: Draw Ray Diagram for Concave Mirror

Depict the principal axis, the focal point, and the center of curvature on the diagram. Position the object 100 cm from the mirror. Draw a ray parallel to the principal axis, reflecting through the focal point, and a ray through the focal point reflecting parallel to the principal axis. Extend rays to find the image's position and characteristics.

Step 3: Analyze Image Characteristics from Ray Diagram

Observing the ray diagram, you can determine: (1) The image is real if rays converge, (2) the image is inverted relative to the object, and (3) the image size is reduced since the image is between the mirror and the focal point.

Step 4: Use Mirror Equation to Calculate Image Distance

Apply the mirror equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( d_o = 100 \mathrm{~cm} \) is the object distance, and \( f = 40 \mathrm{~cm} \) is the focal length. Rearrange for \( d_i \): \( \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{40} - \frac{1}{100} \). Calculating gives \( \frac{1}{d_i} = \frac{5 - 2}{200} = \frac{3}{200} \), so \( d_i = \frac{200}{3} \approx 66.67 \mathrm{~cm} \).

Step 5: Compute Magnification

The magnification \( m \) is given by \( m = -\frac{d_i}{d_o} \). Using \( d_i \approx 66.67 \mathrm{~cm} \) and \( d_o = 100 \mathrm{~cm} \), calculate \( m = -\frac{66.67}{100} = -0.667 \). This indicates that the image is inverted and reduced in size compared to the object.

Key Concepts

Ray Diagram

A ray diagram for a concave mirror is a powerful tool to visualize how light rays interact with a mirror to form an image. To construct a ray diagram, begin by drawing the principal axis, the straight line that passes through the center of the mirror surface and the focal point. Next, identify key points like the center of curvature and the focal point.
Position the object at its proper place on the principal axis, in this case, 100 cm from the mirror. From the top of the object, draw:

• A ray parallel to the principal axis, which will reflect through the focal point.
• A ray passing through the focal point, which will reflect parallel to the principal axis.

By extending these rays on the reflected side, the point where they converge indicates the position of the image. In this example, the rays will converge on the same side of the mirror as the object, showing the image as real and inverted. This visualization also helps to see whether the image is magnified or reduced.

Focal Length

The focal length is an essential characteristic of a concave mirror that determines its ability to converge or diverge rays. It is derived from the radius of curvature, which is the radius of the sphere that the mirror surface is part of. For concave mirrors, the focal length \( f \) can be calculated using the formula:\[ f = \frac{R}{2} \]where \( R \) is the radius of curvature.
In this situation, where \( R = 80 \mathrm{~cm} \), the focal length becomes \( f = 40 \mathrm{~cm} \). This value informs us of the distance from the mirror at which parallel light rays will converge to focus, crucial for predicting image formation.

Mirror Equation

The mirror equation links the object distance \( d_o \), the image distance \( d_i \), and the focal length \( f \). The formula is given by:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]This equation allows us to calculate unknown values when others are known. To find the image distance \( d_i \), rearrange it to:\[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \]Substituting the values \( f = 40 \mathrm{~cm} \) and \( d_o = 100 \mathrm{~cm} \) results in:\[ \frac{1}{d_i} = \frac{1}{40} - \frac{1}{100} = \frac{3}{200} \]Therefore, \( d_i \approx 66.67 \mathrm{~cm} \). The image is found closer to the mirror than the object, consistent with observations that the image is real, inverted, and reduced in size.

Lateral Magnification

Lateral magnification provides insight into how the size of an image compares to the size of the object. It is calculated using the relationship:\[ m = -\frac{d_i}{d_o} \]This formula includes a negative sign to demonstrate that images formed by concave mirrors are inverted.

• For our case, with \( d_i = 66.67 \mathrm{~cm} \) and \( d_o = 100 \mathrm{~cm} \), the magnification \( m \) is:

\[ m = -\frac{66.67}{100} = -0.667 \]This result tells us that the image is inverted and about 66.7% the height of the object, indicating it is reduced in size. Understanding magnification is key for applications where the size of the image matters.