Practice worksheet for drawing ray diagrams and determining image characteristics for convex lenses.

Let's solve each problem step by step using the principles of convex lens imaging. We’ll use the lens formula and ray diagram rules to determine the image characteristics: location, direction (orientation), size, and type (real or virtual).

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🔍 Key Concepts:

- Focal length (f) = 15.0 cm for all lenses (convex)
- Object distance (do) is given in each case
- Use lens formula:
$$
\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}
$$
where:
- $ f $ = focal length
- $ d_o $ = object distance
- $ d_i $ = image distance

We'll also use ray tracing rules:
1. A ray parallel to the principal axis passes through the focal point after refraction.
2. A ray through the center of the lens continues straight.
3. A ray through the focal point emerges parallel to the axis.

Image properties depend on object position relative to F and 2F:
- Beyond 2F: Real, inverted, reduced
- At 2F: Real, inverted, same size
- Between F and 2F: Real, inverted, enlarged
- At F: No image (rays parallel)
- Between F and lens: Virtual, upright, enlarged

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## ✔ Problem 1:
> A 15.0 cm object is placed 60.0 cm from a convex lens with focal length 15.0 cm.

Given:

- $ d_o = 60.0 $ cm
- $ f = 15.0 $ cm

Step 1: Use lens formula

$$
\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{15} - \frac{1}{60}
= \frac{4 - 1}{60} = \frac{3}{60} = \frac{1}{20}
\Rightarrow d_i = 20.0 \text{ cm}
$$

So image is 20.0 cm on the opposite side of the lens.

Step 2: Analyze:

- Object is at 60 cm, which is beyond 2F (since 2F = 30 cm).
- So image should be:
- Real (on opposite side)
- Inverted
- Reduced (smaller than object)

Magnification:

$$
m = -\frac{d_i}{d_o} = -\frac{20}{60} = -\frac{1}{3}
$$
Negative → inverted; magnitude < 1 → reduced

Fill in table:

| Location | Between F and 2F (on other side) |
|----------------|----------------------------------|
| Direction | Inverted |
| Size | Reduced |
| Type | Real |

✔ Answer: Image is real, inverted, reduced, located between F and 2F on the opposite side.

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## ✔ Problem 2:
> A 15.0 cm object is placed 30.0 cm from a convex lens with focal length 15.0 cm.

Given:

- $ d_o = 30.0 $ cm
- $ f = 15.0 $ cm

Step 1: Lens formula

$$
\frac{1}{d_i} = \frac{1}{15} - \frac{1}{30} = \frac{2 - 1}{30} = \frac{1}{30}
\Rightarrow d_i = 30.0 \text{ cm}
$$

Step 2: Analyze:

- Object is at 30 cm, which is exactly at 2F (since 2F = 30 cm)
- Image is also at 2F on the other side
- Image is:
- Real
- Inverted
- Same size as object

Magnification:
$$
m = -\frac{30}{30} = -1
$$
→ Inverted, same size

Fill in table:

| Location | At 2F (on other side) |
|----------------|----------------------------------|
| Direction | Inverted |
| Size | Same size |
| Type | Real |

✔ Answer: Image is real, inverted, same size, located at 2F on the opposite side.

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## ✔ Problem 3:
> A 15.0 cm object is placed 16.0 cm from a convex lens with focal length 15.0 cm.

Given:

- $ d_o = 16.0 $ cm
- $ f = 15.0 $ cm

Step 1: Lens formula

$$
\frac{1}{d_i} = \frac{1}{15} - \frac{1}{16} = \frac{16 - 15}{240} = \frac{1}{240}
\Rightarrow d_i = 240 \text{ cm}
$$

Wait — that’s positive? Let’s double-check:

$$
\frac{1}{d_i} = \frac{1}{15} - \frac{1}{16} = \frac{16 - 15}{240} = \frac{1}{240} \Rightarrow d_i = 240 \text{ cm}
$$

Yes, positive → real image?

But wait! Object is between F and 2F? Let’s check:

- F = 15 cm, 2F = 30 cm
- Object at 16 cm → between F and 2F

So we expect a real, inverted, enlarged image beyond 2F.

And yes, $ d_i = 240 $ cm → far beyond 2F.

Magnification:

$$
m = -\frac{d_i}{d_o} = -\frac{240}{16} = -15
$$
→ Highly enlarged, inverted

Fill in table:

| Location | Beyond 2F (on other side) |
|----------------|----------------------------------|
| Direction | Inverted |
| Size | Enlarged |
| Type | Real |

✔ Answer: Image is real, inverted, enlarged, located beyond 2F on the opposite side.

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## 📐 Ray Diagrams (Summary):

For each case, draw three rays from the top of the object:

1. Parallel to axis → refracts through F
2. Through center → straight line
3. Through F → refracts parallel

The intersection point gives the image.

For Problem 1 (do = 60 cm):

- Rays converge between F and 2F → smaller inverted image

For Problem 2 (do = 30 cm):

- Rays converge at 2F → same size, inverted

For Problem 3 (do = 16 cm):

- Rays converge beyond 2F → larger, inverted image

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## ✔ Final Answers:

Problem 1:

| Location | Between F and 2F (on other side) |
|----------------|----------------------------------|
| Direction | Inverted |
| Size | Reduced |
| Type | Real |

Problem 2:

| Location | At 2F (on other side) |
|----------------|----------------------------------|
| Direction | Inverted |
| Size | Same size |
| Type | Real |

Problem 3:

| Location | Beyond 2F (on other side) |
|----------------|----------------------------------|
| Direction | Inverted |
| Size | Enlarged |
| Type | Real |

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Let me know if you'd like hand-drawn ray diagrams sketched out or explained further!