E14. The objective lens of a microscope has a focal length of 0.6 cm . An object on the microscope slide is placed at a distance of 0.9 cm from the lens. a. At what distance from the lens is the image formed by the objective lens? b. What is the magnification of this image?

Let's solve each part of the problem step by step using the lens formula and magnification principles. ### Given: - **Focal length of the objective lens (\(f\))**: 0.6 cm - **Object distance from the lens (\(u\))**: 0.9 cm ### a. **Finding the Image Distance (\(v\))** We use the **lens formula**: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \] Plugging in the known values: \[ \frac{1}{0.6} = \frac{1}{0.9} + \frac{1}{v} \] \[ \frac{5}{3} = \frac{10}{9} + \frac{1}{v} \] Subtract \(\frac{10}{9}\) from both sides to solve for \(\frac{1}{v}\): \[ \frac{5}{3} - \frac{10}{9} = \frac{1}{v} \] \[ \frac{15}{9} - \frac{10}{9} = \frac{5}{9} = \frac{1}{v} \] Taking the reciprocal to find \(v\): \[ v = \frac{9}{5} = 1.8 \text{ cm} \] **Answer to part a:** The image is formed **1.8 cm** from the lens. ### b. **Calculating the Magnification (\(m\))** Magnification is given by the formula: \[ m = \frac{v}{u} \] Using the values from part a: \[ m = \frac{1.8}{0.9} = 2 \] **Interpretation:** - **Positive magnification** indicates that the image is **real and inverted**. - The **magnitude of magnification** is **2**, meaning the image is **twice** the size of the object. **Answer to part b:** The magnification of the image is **+2**. --- **Summary:** a. The image is formed at **1.8 cm** from the lens. b. The magnification of the image is **+2**.