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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?

Ask by Sherman Ball. in the United States
Jan 23,2025

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Answer

a. The image is formed 1.8 cm from the lens. b. The magnification of the image is 2×.

Solution

Let's solve the problem step by step using the thin lens formula and the magnification formula. ### Given: - **Focal length of the objective lens (f):** 0.6 cm - **Object distance (u):** 0.9 cm ### a. **Finding the Image Distance (v)** The **thin lens formula** is given by: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] Rearranging to solve for \( v \): \[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \] Substitute the given values: \[ \frac{1}{v} = \frac{1}{0.6\ \text{cm}} - \frac{1}{0.9\ \text{cm}} = \frac{5}{3} - \frac{10}{9} = \frac{15}{9} - \frac{10}{9} = \frac{5}{9} \] \[ v = \frac{9}{5} = 1.8\ \text{cm} \] **Image Distance, \( v = 1.8\ \text{cm} \)** ### b. **Calculating the Magnification (m)** The **magnification formula** is: \[ m = \frac{v}{u} \] Substitute the known values: \[ m = \frac{1.8\ \text{cm}}{0.9\ \text{cm}} = 2 \] **Magnification, \( m = 2 \) (or \( 2 \times \))** ### **Summary:** - **a.** The image is formed **1.8 cm** from the lens. - **b.** The magnification of the image is **2×**.

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To find the distance at which the image is formed by the objective lens, we can use the lens formula: \[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\] Where: - \(f\) is the focal length (0.6 cm), - \(v\) is the image distance, - \(u\) is the object distance (-0.9 cm, negative as per the convention). Rearranging the formula gives: \[\frac{1}{v} = \frac{1}{f} + \frac{1}{u}\] Substituting the values: \[\frac{1}{v} = \frac{1}{0.6} + \frac{1}{-0.9}\] Calculating the right side: \[\frac{1}{v} = \frac{1.6667}{1} - \frac{1.1111}{1} = \frac{0.5556}{1}\] Thus: \[v \approx 1.8 \text{ cm}\] Now, for magnification, it is given by: \[M = -\frac{v}{u}\] Substituting the known values: \[M = -\frac{1.8}{-0.9} = 2\] Therefore, the image is formed at a distance of approximately 1.8 cm, and the magnification of this image is 2.

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