Challenge Question! A person uses a converging lens that has a focal length of $6.39 \mathrm{~cm}$ to inspect a gem. The lens forms a virtual image $15.9 \mathrm{~cm}$ away. What is the magnification? Answer: Check

Step 1 ：Given the focal length (f) and the distance of the image (v), we can use the lens formula to find the distance of the object (u):

Step 2 ：\(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)

Step 3 ：\(\frac{1}{u} = \frac{1}{v} - \frac{1}{f}\)

Step 4 ：Plug in the given values (f = 6.39 cm and v = 15.9 cm) and solve for u:

Step 5 ：\(\frac{1}{u} = \frac{1}{15.9} - \frac{1}{6.39}\)

Step 6 ：\(u \approx -10.68 \mathrm{~cm}\)

Step 7 ：Now that we have the distance of the object (u), we can calculate the magnification (M) using the magnification formula:

Step 8 ：\(M = -\frac{v}{u}\)

Step 9 ：Plug in the values (v = 15.9 cm and u = -10.68 cm) and solve for M:

Step 10 ：\(M \approx -\frac{15.9}{-10.68}\)

Step 11 ：\(\boxed{M \approx 1.49}\)