Let's solve this problem step by step using the thin-lens equation and the given information.

Given Data:

1. Focal length of the first lens, $f_1 = 10.0 \, \text{cm}$
2. Focal length of the second lens, $f_2 = 20.0 \, \text{cm}$
3. Distance between the two lenses = $20.0 \, \text{cm}$
4. Distance of the object from lens 1, $p_1 = 30.0 \, \text{cm}$

We need to find:

1. The position of the final image.
2. The total magnification.

Step 1: Image formed by lens 1

We use the thin-lens equation for lens 1: $\frac{1}{f_1} = \frac{1}{p_1} + \frac{1}{q_1}$ Rearranging for $q_1$, the image distance for lens 1: $\frac{1}{q_1} = \frac{1}{f_1} - \frac{1}{p_1}$ Substitute the values: $\frac{1}{q_1} = \frac{1}{10.0} - \frac{1}{30.0}$ $\frac{1}{q_1} = 0.1 - 0.0333 = 0.0667$ $q_1 = \frac{1}{0.0667} \approx 15.0 \, \text{cm}$

So, the image formed by lens 1 is located 15.0 cm to the right of lens 1.

Step 2: Object distance for lens 2

The distance between the two lenses is 20.0 cm. The image formed by lens 1 acts as the object for lens 2. The object distance for lens 2, $p_2$, is: $p_2 = 20.0 \, \text{cm} - 15.0 \, \text{cm} = 5.0 \, \text{cm}$

Step 3: Image formed by lens 2

Now, we use the thin-lens equation for lens 2: $\frac{1}{f_2} = \frac{1}{p_2} + \frac{1}{q_2}$ Rearranging for $q_2$, the image distance for lens 2: $\frac{1}{q_2} = \frac{1}{f_2} - \frac{1}{p_2}$ Substitute the values: $\frac{1}{q_2} = \frac{1}{20.0} - \frac{1}{5.0}$ $\frac{1}{q_2} = 0.05 - 0.2 = -0.15$ $q_2 = \frac{1}{-0.15} \approx -6.67 \, \text{cm}$

The negative sign indicates that the final image is virtual and located 6.67 cm to the left of lens 2.

Step 4: Total magnification

The total magnification is the product of the magnifications of the two lenses: $M = M_1 \cdot M_2$ For lens 1, the magnification is: $M_1 = -\frac{q_1}{p_1} = -\frac{15.0}{30.0} = -0.5$ For lens 2, the magnification is: $M_2 = -\frac{q_2}{p_2} = -\frac{-6.67}{5.0} = 1.33$ The total magnification is: $M = M_1 \cdot M_2 = (-0.5) \cdot (1.33) = -0.665$

The negative sign indicates that the final image is inverted relative to the original object.

Final Results:

1. The position of the final image is 6.67 cm to the left of lens 2.
2. The total magnification is -0.665.

Would you like further clarifications or assistance with related problems?

Related Questions:

1. How does the focal length affect the magnification of a lens system?
2. What happens if the lenses are moved closer together in this system?
3. How would the final image location change if the object distance for lens 1 was reduced?
4. Can you explain why the final image is virtual in this example?
5. What is the significance of the total magnification being less than 1?

Tip:

Always draw ray diagrams alongside solving lens problems to visualize the image formation and verify your calculations.