Edits to “Solution”, “Answer”
en/13.1.19.md
+8 −2
| ### Statement | |||
| $13.1.19.$ [Insert the problem statement] | |||
| @@ -5,7 +5,13 @@Statement | |||
| ### Solution | |||
| Knowing, that the focal length of a spherical Lens is $$\frac{R}{2}$$ | |||
| − | |||
| + | We can use the formula: | ||
| + | $$\frac{1}{f} = \frac{1}{b} + \frac{1}{g}$$ | ||
| + | to get the position of the Image. In this case $g = -R$ and $f = -\frac{R}{2}$(minus because it is a concave lens).Putting these values in the formula: | ||
| + | $$-\frac{2}{R} + \frac{1}{R} = -\frac{1}{R} = \frac{1}{b} \Leftrightarrow b = -R $$ | ||
| + | This tells us that the image is at the exact same position as the object. The formula for the magnification is: | ||
| + | $$M = -\frac{b}{g}$$ | ||
| + | in this case, $b = g$ so $M=-1$, that means, the image is just a flipped version of the object(mirrored along the x-axis) | ||
| #### Answer | |||
| − | [ | ||
| + | [The image is identical to the object(same size and position) but just flipped along the x-axis] | ||
| ### Statement | ### Statement | ||
| $13.1.19.$ [Insert the problem statement] | $13.1.19.$ [Insert the problem statement] | ||
| @@ -5,7 +5,13 @@Statement | |||
| ### Solution | ### Solution | ||
| Knowing, that the focal length of a spherical Lens is $$\frac{R}{2}$$ | Knowing, that the focal length of a spherical Lens is $$\frac{R}{2}$$ | ||
| We can use the formula: | |||
| $$\frac{1}{f} = \frac{1}{b} + \frac{1}{g}$$ | |||
| to get the position of the Image. In this case $g = -R$ and $f = -\frac{R}{2}$(minus because it is a concave lens).Putting these values in the formula: | |||
| $$-\frac{2}{R} + \frac{1}{R} = -\frac{1}{R} = \frac{1}{b} \Leftrightarrow b = -R $$ | |||
| This tells us that the image is at the exact same position as the object. The formula for the magnification is: | |||
| $$M = -\frac{b}{g}$$ | |||
| in this case, $b = g$ so $M=-1$, that means, the image is just a flipped version of the object(mirrored along the x-axis) | |||
| #### Answer | #### Answer | ||
| [ |
[The image is identical to the object(same size and position) but just flipped along the x-axis] | ||