For each of the following problems, draw a ray diagram to help you determine the image location and approximate size. Then, use the mirror equations to solve quantitatively. Do these on your own and then come back here to check your work.
| Problem | Picture | Given/Find | Equation | Solution |
|---|---|---|---|---|
| A converging mirror has a focal length of 40 cm. If an object is placed 50 cm in front of the mirror describe (qualitatively and quantitatively) the characteristics of the image formed in terms of location, type, and magnification. |
|
\(\mathsf{ f = 40 \text{ cm} }\) \(\mathsf{ s_o = 50 \text{ cm} }\) \(\mathsf{ s_i = ? }\) \(\mathsf{ M = ? }\) |
\(\mathsf{ \frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i} }\) \(\mathsf{ M = \frac{h_i}{h_o} = \frac{-s_i}{s_o} }\) |
\(\mathsf{ \frac{1}{40 \text{ cm}} = \frac{1}{50 \text{ cm}} + \frac{1}{s_i} }\) \(\mathsf{ \frac{1}{s_i} = \frac{1}{40 \text{ cm}} - \frac{1}{50 \text{ cm}}}\) \(\mathsf{ \frac{1}{s_i} = \frac{1}{200} }\) \(\mathsf{ s_i = 200 \text{ cm} }\) \(\mathsf{ M = \frac{-200 \text{ cm}}{50 \text{ cm}} = -4 }\) The image is an inverted, real image located 200 cm in front of the mirror. It is four times the size of the original object. |
| A diverging mirror has a focal length of 25 cm. If an object is placed 15 cm in from of the mirror describe (qualitatively and quantitatively) the characteristics of the image formed in terms of location, type, and magnification. |
|
\(\mathsf{ f = -25 \text{ cm} }\) \(\mathsf{ s_o = 15 \text{ cm} }\) \(\mathsf{ s_i = ? }\) \(\mathsf{ M = ? }\) |
\(\mathsf{ \frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i} }\) \(\mathsf{ M = \frac{h_i}{h_o} = \frac{-s_i}{s_o} }\) |
\(\mathsf{ \frac{1}{-25 \text{ cm}} = \frac{1}{15 \text{ cm}} + \frac{1}{s_i} }\) \(\mathsf{ \frac{1}{s_i} = \frac{1}{-25 \text{ cm}} - \frac{1}{15 \text{ cm}}}\) \(\mathsf{ \frac{1}{s_i} = -\frac{8}{75} }\) \(\mathsf{ s_i = -9.4 \text{ cm} }\) \(\mathsf{ M = \frac{-(-9.4 \text{ cm})}{15 \text{ cm}} = 0.63 }\) The image is an upright, virtual image located 9.4 cm behind the mirror. It is 0.63 times the size of the original object. |
| A converging mirror has a focal length of 40 cm. If an object is placed at the focal point in front of the mirror describe (qualitatively and quantitatively) the characteristics of the image formed in terms of location, type, and magnification. |
|
\(\mathsf{ f = 40 \text{ cm} }\) \(\mathsf{ s_o = 40 \text{ cm} }\) \(\mathsf{ s_i = ? }\) \(\mathsf{ M = ? }\) |
\(\mathsf{ \frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i} }\) \(\mathsf{ M = \frac{h_i}{h_o} = \frac{-s_i}{s_o} }\) |
\(\mathsf{ \frac{1}{40 \text{ cm}} = \frac{1}{40 \text{ cm}} + \frac{1}{s_i} }\) \(\mathsf{ \frac{1}{s_i} = \frac{1}{40 \text{ cm}} - \frac{1}{40 \text{ cm}}}\) \(\mathsf{ \frac{1}{s_i} = 0 }\) \(\mathsf{ s_i = \text{undefined} }\) Since the image distance is undefined, there is no image formed when the object is placed at the focal point of a converging mirror. |
| A diverging mirror has a focal length of 25 cm. If an object is placed 50 cm in front of the mirror describe (qualitatively and quantitatively) the characteristics of the image formed in terms of location, type, and magnification. |
|
\(\mathsf{ f = -25 \text{ cm} }\) \(\mathsf{ s_o = 50 \text{ cm} }\) \(\mathsf{ s_i = ? }\) \(\mathsf{ M = ? }\) |
\(\mathsf{ \frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i} }\) \(\mathsf{ M = \frac{h_i}{h_o} = \frac{-s_i}{s_o} }\) |
\(\mathsf{ \frac{1}{-25 \text{ cm}} = \frac{1}{50 \text{ cm}} + \frac{1}{s_i} }\) \(\mathsf{ \frac{1}{s_i} = \frac{1}{-25 \text{ cm}} - \frac{1}{50 \text{ cm}}}\) \(\mathsf{ \frac{1}{s_i} = -\frac{3}{50} }\) \(\mathsf{ s_i = -16.7 \text{ cm} }\) \(\mathsf{ M = \frac{-(-16.7 \text{ cm})}{50 \text{ cm}} = 0.33 }\) The image is an upright, virtual image located 16.7 cm behind the mirror. It is 0.33 times the size of the original object. |
