14-09,10 - Worksheet - Mirrors - Free Printable

Problem Analysis:

The worksheet involves solving problems related to spherical mirrors, specifically concave mirrors. The key equations for solving these problems are:

1. Mirror Formula:
\[
\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}
\]
where:
- \( d_o \) is the object distance,
- \( d_i \) is the image distance,
- \( f \) is the focal length.

2. Magnification Formula:
\[
\frac{h_i}{h_o} = -\frac{d_i}{d_o}
\]
where:
- \( h_i \) is the height of the image,
- \( h_o \) is the height of the object.

3. Sign Conventions:
- For a concave mirror:
- \( f \) is positive.
- \( d_o \) is positive if the object is in front of the mirror.
- \( d_i \) is positive for a real image and negative for a virtual image.
- \( h_i \) is positive for an upright image and negative for an inverted image.

Problem 1:

#### Given:
- Focal length, \( f = 18 \, \text{cm} \)
- Object distance, \( d_o = 58 \, \text{cm} \)
- Height of the object, \( h_o = 12 \, \text{cm} \)

#### To Find:
1. Image distance, \( d_i \)
2. Height of the image, \( h_i \)
3. Nature of the image (real/virtual, erect/inverted)

#### Solution:
1. Using the Mirror Formula:
\[
\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}
\]
Substitute the given values:
\[
\frac{1}{58} + \frac{1}{d_i} = \frac{1}{18}
\]
Solve for \( \frac{1}{d_i} \):
\[
\frac{1}{d_i} = \frac{1}{18} - \frac{1}{58}
\]
Find a common denominator:
\[
\frac{1}{d_i} = \frac{58 - 18}{18 \times 58} = \frac{40}{1044} = \frac{20}{522} = \frac{10}{261}
\]
Therefore:
\[
d_i = \frac{261}{10} = 26.1 \, \text{cm}
\]

2. Using the Magnification Formula:
\[
\frac{h_i}{h_o} = -\frac{d_i}{d_o}
\]
Substitute the known values:
\[
\frac{h_i}{12} = -\frac{26.1}{58}
\]
Solve for \( h_i \):
\[
h_i = 12 \times \left( -\frac{26.1}{58} \right) = 12 \times (-0.45) = -5.4 \, \text{cm}
\]

3. Nature of the Image:
- Since \( d_i \) is positive, the image is real.
- Since \( h_i \) is negative, the image is inverted.

#### Final Answer for Problem 1:
\[
\boxed{d_i = 26.1 \, \text{cm}, \, h_i = -5.4 \, \text{cm}, \, \text{real, inverted}}
\]

Problem 2:

#### Given:
- Focal length, \( f = 18 \, \text{cm} \)
- Object distance, \( d_o = 36 \, \text{cm} \)

#### To Find:
1. Image distance, \( d_i \)
2. Height of the image, \( h_i \)
3. Nature of the image (real/virtual, erect/inverted)

#### Solution:
1. Using the Mirror Formula:
\[
\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}
\]
Substitute the given values:
\[
\frac{1}{36} + \frac{1}{d_i} = \frac{1}{18}
\]
Solve for \( \frac{1}{d_i} \):
\[
\frac{1}{d_i} = \frac{1}{18} - \frac{1}{36}
\]
Find a common denominator:
\[
\frac{1}{d_i} = \frac{2 - 1}{36} = \frac{1}{36}
\]
Therefore:
\[
d_i = 36 \, \text{cm}
\]

2. Using the Magnification Formula:
\[
\frac{h_i}{h_o} = -\frac{d_i}{d_o}
\]
Substitute the known values:
\[
\frac{h_i}{12} = -\frac{36}{36} = -1
\]
Solve for \( h_i \):
\[
h_i = 12 \times (-1) = -12 \, \text{cm}
\]

3. Nature of the Image:
- Since \( d_i \) is positive, the image is real.
- Since \( h_i \) is negative, the image is inverted.

#### Final Answer for Problem 2:
\[
\boxed{d_i = 36 \, \text{cm}, \, h_i = -12 \, \text{cm}, \, \text{real, inverted}}
\]

Problem 3:

#### Given:
- Focal length, \( f = 18 \, \text{cm} \)
- Object distance, \( d_o = 32 \, \text{cm} \)

#### To Find:
1. Image distance, \( d_i \)
2. Height of the image, \( h_i \)
3. Nature of the image (real/virtual, erect/inverted)

#### Solution:
1. Using the Mirror Formula:
\[
\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}
\]
Substitute the given values:
\[
\frac{1}{32} + \frac{1}{d_i} = \frac{1}{18}
\]
Solve for \( \frac{1}{d_i} \):
\[
\frac{1}{d_i} = \frac{1}{18} - \frac{1}{32}
\]
Find a common denominator:
\[
\frac{1}{d_i} = \frac{32 - 18}{18 \times 32} = \frac{14}{576} = \frac{7}{288}
\]
Therefore:
\[
d_i = \frac{288}{7} \approx 41.14 \, \text{cm}
\]

2. Using the Magnification Formula:
\[
\frac{h_i}{h_o} = -\frac{d_i}{d_o}
\]
Substitute the known values:
\[
\frac{h_i}{12} = -\frac{41.14}{32}
\]
Solve for \( h_i \):
\[
h_i = 12 \times \left( -\frac{41.14}{32} \right) = 12 \times (-1.2856) \approx -15.43 \, \text{cm}
\]

3. Nature of the Image:
- Since \( d_i \) is positive, the image is real.
- Since \( h_i \) is negative, the image is inverted.

#### Final Answer for Problem 3:
\[
\boxed{d_i \approx 41.14 \, \text{cm}, \, h_i \approx -15.43 \, \text{cm}, \, \text{real, inverted}}
\]

Summary of Answers:

1. \( d_i = 26.1 \, \text{cm}, \, h_i = -5.4 \, \text{cm}, \, \text{real, inverted} \)
2. \( d_i = 36 \, \text{cm}, \, h_i = -12 \, \text{cm}, \, \text{real, inverted} \)
3. \( d_i \approx 41.14 \, \text{cm}, \, h_i \approx -15.43 \, \text{cm}, \, \text{real, inverted} \)