The image contains the lens equation, a common formula in optics:

$\frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i}$

Where:

• $f$ is the focal length of the lens,
• $s_o$ is the object distance (distance from the object to the lens),
• $s_i$ is the image distance (distance from the image to the lens).

It seems the equation is being solved for $s_i$ (image distance). We can rearrange the equation to solve for $s_i$:

$\frac{1}{s_i} = \frac{1}{f} - \frac{1}{s_o}$

Thus, the formula for $s_i$ becomes:

$s_i = \frac{1}{\left( \frac{1}{f} - \frac{1}{s_o} \right)}$

Would you like to solve a specific problem using this formula, or would you like a further explanation?

5 Related Questions:

1. What is the focal length $f$ in a typical lens problem?
2. How does changing the object distance $s_o$ affect the image distance $s_i$?
3. Can this formula be used for concave and convex lenses?
4. How does magnification relate to the object and image distances?
5. What are the conditions for the image to be real or virtual in a lens?

Tip:

Always keep units consistent when using the lens equation (e.g., all distances in meters or centimeters).