Problem 10 - Physics

Question

At illustrates the concepts pertinent to this problem. A \(2.0-\mathrm{cm}-\) high object is situated \(15.0 \mathrm{~cm}\) in front of a concave mirror that has a radius of curvature of \(10.0 \mathrm{~cm}\). Using a ray diagram drawn to scale, measure (a) the location and (b) the height of the image. The mirror must be drawn to scale.

Step-by-Step Solution

Answer

The image is located 7.5 cm in front of the mirror and is 1.0 cm tall, inverted.

Step 1: Determine the Mirror's Focal Length

The radius of curvature \( R \) of the mirror is \( 10.0 \mathrm{~cm} \). The focal length \( f \) of a concave mirror is half the radius of curvature. \[ f = \frac{R}{2} = \frac{10.0 \mathrm{~cm}}{2} = 5.0 \mathrm{~cm} \]

Step 2: Apply Mirror Equation

To find the image distance, use the mirror equation:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]where \( d_o = 15.0 \mathrm{~cm} \) is the object distance.\[ \frac{1}{5.0} = \frac{1}{15.0} + \frac{1}{d_i} \]Solving for \( \frac{1}{d_i} \), we have:\[ \frac{1}{d_i} = \frac{1}{5.0} - \frac{1}{15.0} = \frac{3 - 1}{15} = \frac{2}{15} \]Thus, the image distance \( d_i \) is:\[ d_i = \frac{15}{2} = 7.5 \mathrm{~cm} \]

Step 3: Calculate Image Height Using Magnification Equation

The magnification \( m \) is given by the formula:\[ m = -\frac{d_i}{d_o} = -\frac{7.5}{15.0} = -0.5 \]The image height \( h_i \) is related to the object height \( h_o = 2.0 \mathrm{~cm} \):\[ h_i = m \times h_o = -0.5 \times 2.0 = -1.0 \mathrm{~cm} \]The negative sign indicates that the image is inverted.

Step 4: Interpret the Ray Diagram

Using the calculated image distance and magnification, draw a ray diagram to confirm image position and height: - Draw principal axis, mirror with a vertex at the origin and focus at 5.0 cm. - Mark object 15.0 cm from the mirror and 2.0 cm high. - Light rays from the top of the object passing through the focus (after reflection) and parallel to the principal will converge at 7.5 cm, forming an inverted image. - Verify the image height measures -1.0 cm below the principal axis.

Key Concepts

Focal Length

The focal length of a concave mirror is a fundamental property to understand. It is half the radius of curvature of the mirror, denoted as \( f \). This relationship is expressed mathematically as \( f = \frac{R}{2} \), where \( R \) is the radius of curvature. In our exercise, \( R \) is given as \( 10.0 \text{ cm} \). Hence, the focal length is \( \frac{10.0}{2} = 5.0 \text{ cm} \). This means that the focal point is located 5 cm from the mirror along the principal axis. Understanding focal length helps in predicting where light rays will converge after reflection.

Mirror Equation

The mirror equation is a crucial tool used to relate focal length, object distance, and image distance. It is derived from the geometric principles of optics and is expressed as \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Here, \( d_o \) is the object distance and \( d_i \) is the image distance.

• For this problem, with \( f = 5.0 \text{ cm} \) and \( d_o = 15.0 \text{ cm} \), the equation becomes \( \frac{1}{5.0} = \frac{1}{15.0} + \frac{1}{d_i} \).
• Solving this gives \( d_i = 7.5 \text{ cm} \), indicating where the image is formed along the principal axis.

This equation is invaluable for determining unseen distances in optics problems.

Ray Diagram

A ray diagram offers a visual approach to understanding how concave mirrors form images. It involves drawing light rays emanating from the top of the object reflecting through specific points.

• One ray travels parallel to the principal axis and reflects through the focal point.
• Another ray passes through the focal point and reflects parallel to the principal axis.

Where these reflected rays converge gives the position of the image. In the exercise, the rays intersect at \( 7.5 \text{ cm} \), confirming the calculated image distance and demonstrating the image is inverted.

Image Distance

The image distance \( d_i \) is the distance from the mirror to the image formed by the reflection of the object. Calculating \( d_i \) is achieved using the mirror equation. After inserting known values and solving, the image distance for our example is determined to be \( 7.5 \text{ cm} \). With the image distance known, one can

• understand the real location of the image relative to the mirror,
• and predict the nature of the image (upside down if negative).

The accurate placement of the image on the ray diagram further validates these findings.

Magnification

Magnification describes how much larger or smaller the image is compared to the object. It is calculated using the equation \( m = -\frac{d_i}{d_o} \).

• For this exercise, magnification is \( -\frac{7.5}{15.0} = -0.5 \).
• This negative value indicates that the image is inverted.
• The magnitude of \( 0.5 \) tells us the image is half the height of the object.

In practice, magnification allows us to determine how an image's size compares to its actual object, providing useful insights into its scaling and orientation.