A \(1.80-\mathrm{cm}\) -high object is placed \(20.0 \mathrm{cm}\) in front of a concave mirror with a \(5.00-\mathrm{cm}\) focal length. What is the position of the image? Draw a ray diagram to illustrate.

We are given: Object distance: \(d_o = 20.0\, cm\) Focal length: \(f = 5.00\, cm\) Unknown: Image distance: \(d_i\)

We can rearrange the mirror equation to solve for \(d_i\): $$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}$$ Now, plug in the given values: $$\frac{1}{d_i} = \frac{1}{5.00\, cm} - \frac{1}{20.0\, cm}$$ Calculate the resulting value: $$\frac{1}{d_i} = 0.20\, cm^{-1} - 0.05\, cm^{-1}$$ $$\frac{1}{d_i} = 0.15\, cm^{-1}$$ To find the image distance, take the reciprocal of the result: $$d_i = \frac{1}{0.15\, cm^{-1}}$$ $$d_i = 6.67\, cm$$ So, the position of the image is \(6.67\, cm\) in front of the concave mirror.

To draw a ray diagram follow these steps: 1. Draw a concave mirror and label the center of curvature (C), the focal point (F) halfway between C and the mirror, and the pole (P) at the mirror's vertex. 2. Draw the object in front of the mirror. 3. Draw a ray parallel to the principal axis from the top of the object to the mirror and then through the focal point (F). 4. Draw a ray from the top of the object through the focal point (F) and then parallel to the principal axis to the mirror. 5. The intersection of these two rays will be the location of the top of the image. In this exercise, the image will be formed between the focal point and the center of curvature, and it will be smaller and inverted compared to the object.