Правка разделов «Statement», «Solution», «Answer»
en/13.1.17.md
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| ### Statement | |||
| − | $13.1.17.$ [Insert the problem statement] | ||
| + | $13.1.17.$ The actual image of an object in a concave mirror is three times the size of the object. After the object was moved 80 cm away from the mirror, its image became half the size of the object. Find the focal length of the mirror. | ||
| ### Solution | |||
| Firstly, the magnification of the object is given by:\ | |||
| $k = \frac{s'}{s}$ (1)\ | |||
| where $s'$ is the image-mirror distance and $s$ is the object-mirror distance, with $k$ = 3.\ | |||
| And the relation for these distances is\ | |||
| $\frac{1}{s} +\frac{1}{s'} = \frac{1}{f}$ (2)\ | |||
| When object is moved a distance $d$ away from the mirror, the magnification changes\ | |||
| $m = \frac{s''}{s+d}$ (3)\ | |||
| with $m$ = 0.5.\ | |||
| And the relation for these distances is\ | |||
| @@ -16,12 +16,12 @@Solution | |||
| $\frac{1}{s+d}+\frac{1}{s''}=\frac{1}{f}$ (4)\ | |||
| From (1) and (2), eliminating term $s'$,\ | |||
| $\frac{s}{f} = 1+\frac{1}{k}$ (5)\ | |||
| − | From (3) and (4), eliminating term $s''$, | ||
| + | From (3) and (4), eliminating term $s''$,\ | ||
| $\frac{s}{f}=1+\frac{1}{m}-\frac{d}{f}$ (6)\ | |||
| Equaling (5) and (6), and separating $f$,\ | |||
| $f = \frac{mkd}{k-m} = 48\;\rm{cm}$ | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $f$ = 48 cm | ||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $13.1.17.$ [Insert the problem statement] | $13.1.17.$ The actual image of an object in a concave mirror is three times the size of the object. After the object was moved 80 cm away from the mirror, its image became half the size of the object. Find the focal length of the mirror. | ||
| ### Solution | ### Solution | ||
| Firstly, the magnification of the object is given by:\ | Firstly, the magnification of the object is given by:\ | ||
| $k = \frac{s'}{s}$ (1)\ | $k = \frac{s'}{s}$ (1)\ | ||
| where $s'$ is the image-mirror distance and $s$ is the object-mirror distance, with $k$ = 3.\ | where $s'$ is the image-mirror distance and $s$ is the object-mirror distance, with $k$ = 3.\ | ||
| And the relation for these distances is\ | And the relation for these distances is\ | ||
| $\frac{1}{s} +\frac{1}{s'} = \frac{1}{f}$ (2)\ | $\frac{1}{s} +\frac{1}{s'} = \frac{1}{f}$ (2)\ | ||
| When object is moved a distance $d$ away from the mirror, the magnification changes\ | When object is moved a distance $d$ away from the mirror, the magnification changes\ | ||
| $m = \frac{s''}{s+d}$ (3)\ | $m = \frac{s''}{s+d}$ (3)\ | ||
| with $m$ = 0.5.\ | with $m$ = 0.5.\ | ||
| And the relation for these distances is\ | And the relation for these distances is\ | ||
| @@ -16,12 +16,12 @@Solution | |||
| $\frac{1}{s+d}+\frac{1}{s''}=\frac{1}{f}$ (4)\ | $\frac{1}{s+d}+\frac{1}{s''}=\frac{1}{f}$ (4)\ | ||
| From (1) and (2), eliminating term $s'$,\ | From (1) and (2), eliminating term $s'$,\ | ||
| $\frac{s}{f} = 1+\frac{1}{k}$ (5)\ | $\frac{s}{f} = 1+\frac{1}{k}$ (5)\ | ||
| From (3) and (4), eliminating term $s''$, | From (3) and (4), eliminating term $s''$,\ | ||
| $\frac{s}{f}=1+\frac{1}{m}-\frac{d}{f}$ (6)\ | $\frac{s}{f}=1+\frac{1}{m}-\frac{d}{f}$ (6)\ | ||
| Equaling (5) and (6), and separating $f$,\ | Equaling (5) and (6), and separating $f$,\ | ||
| $f = \frac{mkd}{k-m} = 48\;\rm{cm}$ | $f = \frac{mkd}{k-m} = 48\;\rm{cm}$ | ||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $f$ = 48 cm | ||