11. An object is 2 cm from a concave mirror that has a focal length of 5 cm . Determine the image location, magnification, real or virtual, and upright or inverted for the object. a. Using the equation:

We are given:  Object distance, do = 2 cm (in front of the mirror)  Focal length, f = 5 cm Step 1. Write down the mirror equation:   1/f = 1/do + 1/di Step 2. Substitute the given values:   1/5 = 1/2 + 1/di Step 3. Solve for 1/di:   1/di = 1/5 – 1/2 Find a common denominator (10):   1/di = (2/10) – (5/10) = –3/10 Step 4. Find di:   di = –10/3 cm ≈ –3.33 cm Interpretation of di:  The negative sign indicates that the image is formed on the same side of the mirror as the object, meaning it is a virtual image. Step 5. Compute the magnification (m) using:   m = –di/do Substitute the values:   m = –(–10/3) / 2 = (10/3) / 2 = 10/6 = 5/3 ≈ 1.67 Interpretation of magnification:  Since m is positive, the image is upright relative to the object.  Since |m| = 5/3 > 1, the image is magnified. Summary of the results:  • Image location: di = –3.33 cm (i.e., the image is 3.33 cm behind the mirror)  • Magnification: m = 5/3 (approximately 1.67 times larger than the object)  • Nature: The image is virtual (because di is negative) and upright (because m is positive).