To solve this lens problem, we use the following information and equations for a converging lens:

Given:

• Object height ($h_o$) = $10.0 \, \text{cm}$
• Image distance ($d_i$) = $60.0 \, \text{cm}$ (real image means $d_i > 0$)
• Focal length ($f$) = $20.0 \, \text{cm}$ (converging lens means $f > 0$)

We are tasked to determine:

1. Object distance ($d_o$)
2. Magnification ($M$)
3. Image height ($h_i$)
4. The size and orientation of the image.

Step 1: Use the Lens Formula to find $d_o$

The lens formula is:

$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

Rearranging for $d_o$:

$\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i}$

Substitute $f = 20.0 \, \text{cm}$ and $d_i = 60.0 \, \text{cm}$:

$\frac{1}{d_o} = \frac{1}{20} - \frac{1}{60}$

$\frac{1}{d_o} = \frac{3}{60} - \frac{1}{60} = \frac{2}{60}$

$d_o = \frac{60}{2} = 30.0 \, \text{cm}$

So, $d_o = 30.0 \, \text{cm}$.

Step 2: Calculate the Magnification ($M$)

The magnification is given by:

$M = -\frac{d_i}{d_o}$

Substitute $d_i = 60.0 \, \text{cm}$ and $d_o = 30.0 \, \text{cm}$:

$M = -\frac{60}{30} = -2$

Thus, $M = -2$.

Step 3: Find the Image Height ($h_i$)

The image height is given by:

$h_i = M \cdot h_o$

Substitute $M = -2$ and $h_o = 10.0 \, \text{cm}$:

$h_i = -2 \cdot 10 = -20.0 \, \text{cm}$

So, the image height $h_i = -20.0 \, \text{cm}$. The negative sign indicates the image is inverted.

Step 4: Size and Orientation of the Image

• Size of the Image: $20.0 \, \text{cm}$ (magnitude of $h_i$).
• Orientation of the Image: Inverted (due to the negative sign of $h_i$).

Final Answers:

• $f = 20.0 \, \text{cm}$

• $d_o = 30.0 \, \text{cm}$

• $d_i = 60.0 \, \text{cm}$

• $M = -2$

• $h_o = 10.0 \, \text{cm}$

• $h_i = -20.0 \, \text{cm}$

• Size of the image: $20.0 \, \text{cm}$

• Orientation of the image: Inverted

Do you need further clarification on any step? Let me know!

Related Questions:

1. How would the results change if the object was placed at twice the focal length?
2. What happens if the focal length of the lens is reduced?
3. What kind of image is formed when the object is inside the focal length?
4. How can you differentiate between real and virtual images using sign conventions?
5. Can you explain the relationship between magnification and object/image distances?

Tip:

When solving lens problems, always ensure to use proper sign conventions: real distances are positive, and inverted images have negative heights.