A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. What is the length of the image?
To find the length of the image, we can use the mirror formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \(f\) is the focal length of the concave mirror,
- \(v\) is the distance of the image from the mirror (which is what we want to calculate),
- \(u\) is the distance of the object from the mirror.
Given that the focal length \(f = -10 \, \text{cm}\), and the distance of the object from the mirror \(u = -20 \, \text{cm}\) (the negative sign indicates that the object is on the same side as the incident light), we can solve for \(v\):
\[ \frac{1}{-10 \, \text{cm}} = \frac{1}{v} - \frac{1}{-20 \, \text{cm}} \]
Simplifying the equation:
\[ -\frac{1}{10 \, \text{cm}} = \frac{1}{v} + \frac{1}{20 \, \text{cm}} \]
Multiplying both sides by \(20v\):
\[ -2v = 20 + v \]
Simplifying the equation further:
\[ -3v = 20 \]
Dividing both sides by -3:
\[ v = -\frac{20}{3} \, \text{cm} \]
The image is formed on the opposite side of the mirror as the object, so the distance of the image is positive. Therefore, the length of the image is:
\[ \text{Length of the image} = 2v = 2 \times \left(-\frac{20}{3}\right) = -\frac{40}{3} \, \text{cm} \]
But the length cannot be negative, so we take the absolute value:
\[ \text{Length of the image} = \left|- \frac{40}{3} \right| = \frac{40}{3} \, \text{cm} \]
Therefore, the length of the image is \( \frac{40}{3} \, \text{cm} \), or approximately \( 13.33 \, \text{cm} \).