Edits to “Statement”, “Solution”, “Answer”
en/13.3.21.md
+5 −5
| @@ -1,15 +1,15 @@ | |||
| ### Statement | |||
| − | $13.3.21.$ [Insert the problem statement] | ||
| + | $13.3.21.$ Two lenses with a focal length of $30$ cm are located at a distance of $15$ cm from each other. Find out at which positions of the object the system gives a valid image. | ||
| ### Solution | |||
| − | In order for the image from the second lens to be a real image, the image from the first lens (which serves as the object for the second lens) has to be farther than $30$ cm (i.e. the focal length) in front of the second lens. This means that the (virtual) image from the first lens has to be farther than $15$ cm in front of the first lens. From the thin lens equation, the position of the object when the image is $15$ cm in front of the first lens is | ||
| + | In order for the image from the second lens to be a real image, the image from the first lens (which serves as the object for the second lens) has to be farther than $30$ cm (i.e. the focal length) in front of the second lens. This means that the (virtual) image from the first lens has to be farther than $15$ cm in front of the first lens, because the two lenses are $15$ cm apart. From the thin lens equation, the position of the object when the image is $15$ cm in front of the first lens is | ||
| − | \[u=\frac{v | ||
| + | \[u=\frac{vf}{v-f}=\frac{-15\cdot30}{-15-30}=-10.\] | ||
| − | Therefore, the object has to be | ||
| + | Therefore, the object has to be less than $10$ cm from the first lens. | ||
| #### Answer | |||
| − | |||
| + | Less than $10$ cm from the first lens. | ||
| @@ -1,15 +1,15 @@ | |||
| ### Statement | ### Statement | ||
| $13.3.21.$ [Insert the problem statement] | $13.3.21.$ Two lenses with a focal length of $30$ cm are located at a distance of $15$ cm from each other. Find out at which positions of the object the system gives a valid image. | ||
| ### Solution | ### Solution | ||
| In order for the image from the second lens to be a real image, the image from the first lens (which serves as the object for the second lens) has to be farther than $30$ cm (i.e. the focal length) in front of the second lens. This means that the (virtual) image from the first lens has to be farther than $15$ cm in front of the first lens. From the thin lens equation, the position of the object when the image is $15$ cm in front of the first lens is | In order for the image from the second lens to be a real image, the image from the first lens (which serves as the object for the second lens) has to be farther than $30$ cm (i.e. the focal length) in front of the second lens. This means that the (virtual) image from the first lens has to be farther than $15$ cm in front of the first lens, because the two lenses are $15$ cm apart. From the thin lens equation, the position of the object when the image is $15$ cm in front of the first lens is | ||
| \[u=\frac{v |
\[u=\frac{vf}{v-f}=\frac{-15\cdot30}{-15-30}=-10.\] | ||
| Therefore, the object has to be |
Therefore, the object has to be less than $10$ cm from the first lens. | ||
| #### Answer | #### Answer | ||
| Less than $10$ cm from the first lens. | |||