Problem 8 - College Physics

Question

\(\cdot\) A concave mirror has a radius of curvature of 34.0 \(\mathrm{cm}\) . (a) What is its focal length? (b) A ladybug 7.50 \(\mathrm{mm}\) tall is located 22.0 \(\mathrm{cm}\) from this mirror along the principal axis. Find the location and height of the image of the insect. (c) If the mirror is immersed in water (of refractive index \(1.33 ),\) what is its focal length?

Step-by-Step Solution

Answer

(a) Focal length is 17.0 cm. (b) Image location is 79.6 cm with height -27.1 mm. (c) Focal length in water is 12.8 cm.

Step 1: Calculate Focal Length Using Radius of Curvature

The focal length \( f \) of a mirror is given by the formula \( f = \frac{R}{2} \), where \( R \) is the radius of curvature of the mirror. Given that \( R = 34.0 \; \text{cm} \), we can calculate the focal length as follows:\[ f = \frac{34.0}{2} = 17.0 \; \text{cm} \].

Step 2: Use Mirror Equation for Image Location

The mirror equation is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( d_o \) is the object distance and \( d_i \) is the image distance. We are given \( d_o = 22.0 \; \text{cm} \) and from Step 1, \( f = 17.0 \; \text{cm} \). Plugging in these values, we get:\[ \frac{1}{17.0} = \frac{1}{22.0} + \frac{1}{d_i} \]Rearrange to find \( d_i \):\[ \frac{1}{d_i} = \frac{1}{17.0} - \frac{1}{22.0} \approx 0.01256 \]Calculate \( d_i \):\[ d_i \approx \frac{1}{0.01256} \approx 79.6 \; \text{cm} \]

Step 3: Determine Image Height Using Magnification

Magnification \( m \) is given by \( m = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \), where \( h_i \) is the image height and \( h_o \) is the object height. We have \( h_o = 7.50 \; \text{mm} \), \( d_i = 79.6 \; \text{cm} \), and \( d_o = 22.0 \; \text{cm} \).\[ m = -\frac{79.6}{22.0} \approx -3.618 \]Calculate \( h_i \):\[ h_i = m \times h_o = -3.618 \times 7.50 \approx -27.1 \; \text{mm} \]

Step 4: Adjust Focal Length for Refraction in Water

The focal length in a medium other than air is given by \( f' = \frac{f}{n} \), where \( n \) is the refractive index of the new medium. Given \( n = 1.33 \), and \( f = 17.0 \; \text{cm} \) from Step 1, the new focal length is:\[ f' = \frac{17.0}{1.33} \approx 12.8 \; \text{cm} \]

Key Concepts

Focal Length Calculation

Understanding the focal length of concave mirrors is crucial for evaluating how light converges to a point. For concave mirrors, the focal length (\( f \)) is calculated using the formula \( f = \frac{R}{2} \), where \( R \) is the radius of curvature. The radius of curvature is the distance between the mirror's surface and the center of its curvature. In our case, with \( R = 34.0 \; \text{cm} \), the calculation proceeds as follows:

\[ f = \frac{34.0}{2} = 17.0 \; \text{cm} \]
This means that the focal point, where rays of light that are parallel to the principal axis converge, is 17.0 cm away from the mirror. This understanding is foundational for further image analysis using concave mirrors.

Mirror Equation

The mirror equation helps us determine the relationship between the object, the image, and the focal distances. It is expressed as: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where:

• \( f \) is the focal length of the mirror.
• \( d_o \) is the distance from the object to the mirror.
• \( d_i \) is the distance from the image to the mirror.

Given \( f = 17.0 \; \text{cm} \) and \( d_o = 22.0 \; \text{cm} \), we can solve for \( d_i \). Rearranging gives:

\[ \frac{1}{d_i} = \frac{1}{17.0} - \frac{1}{22.0} \approx 0.01256 \]
Thus, the image distance \( d_i \) is expressed as:

\[ d_i \approx \frac{1}{0.01256} \approx 79.6 \; \text{cm} \]
This equation is powerful for predicting where an image will form relative to the mirror whenever you know the object distance and focal length.

Image Height Determination

Magnification is the concept that relates the height of the image to the height of the object in optical systems like mirrors. The magnification \( m \) is given by the formula: \( m = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \), where:

• \( h_i \) is the height of the image.
• \( h_o \) is the height of the object.
• \( d_i \) is the image distance.
• \( d_o \) is the object distance.

For the ladybug example, with \( h_o = 7.50 \; \text{mm} \), \( d_i = 79.6 \; \text{cm} \), and \( d_o = 22.0 \; \text{cm} \), the magnification becomes:

\[ m = -\frac{79.6}{22.0} \approx -3.618 \]
Thus, the height of the image \( h_i \) is calculated as:

\[ h_i = m \times h_o = -3.618 \times 7.50 \approx -27.1 \; \text{mm} \]
The negative sign indicates that the image is inverted relative to the object.

Refractive Index Adjustment

When a mirror is placed in a medium other than air, such as water, its focal length adjusts according to the refractive index of the medium. The adjusted focal length \( f' \) is calculated using: \( f' = \frac{f}{n} \), where:

• \( f \) is the original focal length in air.
• \( n \) is the refractive index of the new medium.

For our example, with \( f = 17.0 \; \text{cm} \) and the refractive index of water \( n = 1.33 \), the calculation is:

\[ f' = \frac{17.0}{1.33} \approx 12.8 \; \text{cm} \]
This decrease signifies that light rays converge more quickly in water than in air. It is important to account for the medium when calculating optical properties, ensuring precise results.