Mirror and Magnification Equations
Mirror and Magnification Equations
The mirror equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f).
The equation is stated as follows:
[pic]
The magnification equation relates the ratio of the image distance and object distance to the ratio of the image height (hi) and object height (ho). The magnification equation is stated as follows:
[pic]
These two equations can be combined to yield information about the image distance and image height if the object distance, object height, and focal length are known.
Example Problem #1
A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a concave mirror having a focal length of 15.2 cm. Determine the image distance and the image size. Like all problems in physics, begin by the identification of the known information.
Given:
|ho = 4.0 cm |do = 45.7 cm |f = 15.2 cm |
Next identify the unknown quantities which you wish to solve for.
Required:
|di = ??? |hi = ??? |
Analysis:
1/f = 1/do + 1/di
1/(15.2 cm) = 1/(45.7 cm) + 1/di
0.0658 cm-1 = 0.0219 cm-1 + 1/di
0.0439 cm-1 = 1/di
|di = 22.8 cm |
The final answer is rounded to the third significant digit.
Solution/Paraphrase:
The image is 22.8 cm away.
To determine the image height, the magnification equation is needed.
Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below. hi/ho = - di/do hi /(4.0 cm) = - (22.8 cm)/(45.7 cm) hi = - (4.0 cm) • (22.8 cm)/(45.7 cm)
|hi = -1.99 cm |
The negative values for image height indicate that the image is an inverted image. In the case of the image height, a negative value always indicates an inverted image.
From the
The mirror equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f).
The equation is stated as follows:
[pic]
The magnification equation relates the ratio of the image distance and object distance to the ratio of the image height (hi) and object height (ho). The magnification equation is stated as follows:
[pic]
These two equations can be combined to yield information about the image distance and image height if the object distance, object height, and focal length are known.
Example Problem #1
A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a concave mirror having a focal length of 15.2 cm. Determine the image distance and the image size. Like all problems in physics, begin by the identification of the known information.
Given:
|ho = 4.0 cm |do = 45.7 cm |f = 15.2 cm |
Next identify the unknown quantities which you wish to solve for.
Required:
|di = ??? |hi = ??? |
Analysis:
1/f = 1/do + 1/di
1/(15.2 cm) = 1/(45.7 cm) + 1/di
0.0658 cm-1 = 0.0219 cm-1 + 1/di
0.0439 cm-1 = 1/di
|di = 22.8 cm |
The final answer is rounded to the third significant digit.
Solution/Paraphrase:
The image is 22.8 cm away.
To determine the image height, the magnification equation is needed.
Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below. hi/ho = - di/do hi /(4.0 cm) = - (22.8 cm)/(45.7 cm) hi = - (4.0 cm) • (22.8 cm)/(45.7 cm)
|hi = -1.99 cm |
The negative values for image height indicate that the image is an inverted image. In the case of the image height, a negative value always indicates an inverted image.
From the