91Ó°ÊÓ

To find the equivalent focal length of the lens combination, we'll use the formula for the combined focal length of two thin lenses with focal lengths \(f_1\) and \(f_2\) which are placed very close together: \(1/f_{eq} = 1/f_{1} + 1/f_{2}\) Given that \(f_{1} = 4.00 \mathrm{cm}\) (converging lens) and \(f_{2} = -20.0\mathrm{cm}\) (diverging lens), let's plug in these values in the formula. \(1/f_{eq} = \frac{1}{4.00} - \frac{1}{20.0}\) Calculate \(1/f_{eq}\): \(1/f_{eq} = 0.25 - 0.05 = 0.20 \mathrm{cm^{-1}}\) Now, find out \(f_{eq}\): \(f_{eq} = 1 / 0.20 = 5.00 \mathrm{cm}\) Hence, the equivalent focal length of the lens combination is \(5.00 \mathrm{cm}\).

We will now use the lens formula to find out the position of the image formed by the lens combination. The lens formula is given by: \(1/f_{eq} = 1/u + 1/v\) Where \(u\) is the object distance, \(v\) is the image distance, and \(f_{eq}\) is the equivalent focal length that we found in step 1 (\(5.00 \mathrm{cm}\)). Note that we're given \(u = 2.50 \mathrm{m} = 250 \mathrm{cm}\) (as the object is to the left of the lenses). Now, substitute the given values in the lens formula and solve for \(v\): \(1/5.00 = 1/250 - 1/v\) Solve for \(1/v\): \(1/v = 1/5.00 - 1/250 = 0.20 - 0.004 = 0.196 \mathrm{cm^{-1}}\) Calculate \(v\): \(v = 1 / 0.196 \approx 5.10 \mathrm{cm}\) The image is formed at \(5.10 \mathrm{cm}\) from the lens combination on the same side as the object (since the calculated \(v\) is positive). Now, let's determine if the image is real or virtual. As the lenses are converging (as the equivalent focal length is positive), the image will be real if it's formed on the other side of the lenses and virtual if it's formed on the same side. Since the image is formed on the same side as the object (to the left of the lenses), the image is virtual. In conclusion: (a) The image will be located at \(5.10 \mathrm{cm}\) to the left of the lens combination. (b) The image is virtual.