Diverging lens ray diagrams for various object distances.

It looks like you've uploaded a worksheet titled "Diverging & Converging LENS Worksheet", and it asks you to analyze the image formation by a diverging lens (a concave lens) for different object distances $ s $, using both ray diagrams and the thin lens equation.

Since I can't view images directly, I’ll assume based on your description that the lens in question is a diverging (concave) lens, and the object is placed at various distances:
- $ s = 3f $
- $ s = 2f $
- $ s = f $
- $ s = \frac{1}{2}f $
- $ s = \frac{1}{3}f $

We are to:
1. Draw the three principal rays.
2. Use the thin lens equation to calculate:
- Image height $ y' $
- Image distance $ s' $
- Whether the image is real or virtual
- Whether the image is upright or inverted

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🔍 Key Concepts for Diverging Lenses:

For a diverging (concave) lens:
- Focal length $ f $ is negative: $ f < 0 $
- Always forms virtual, upright, and diminished images regardless of object position.
- The image is always on the same side as the object.

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📐 Thin Lens Equation:

$$
\frac{1}{f} = \frac{1}{s} + \frac{1}{s'}
$$

Where:
- $ s $ = object distance (positive if object is on the side from which light comes)
- $ s' $ = image distance (positive if image is on the opposite side of the lens from the object → real; negative → virtual)
- $ f $ = focal length (negative for diverging lenses)

Magnification:
$$
m = -\frac{s'}{s} = \frac{y'}{y}
$$

Where:
- $ y $ = object height
- $ y' $ = image height
- If $ m > 0 $ → upright
- If $ m < 0 $ → inverted

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Let’s solve each case step-by-step.

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## ✔ Case 1: $ s = 3f $

Assume $ f $ is positive in magnitude, so $ f = |f| $, but since it's a diverging lens, use $ f = -|f| $. Let’s define $ f = -f_0 $, where $ f_0 > 0 $.

So $ s = 3f = 3(-f_0) = -3f_0 $? No — wait!

Wait! In optics convention:
- Object distance $ s $ is always positive when object is on the side from which light comes (left of lens).
- So $ s = 3f $ means $ s = 3|f| $, and $ f $ (focal length) is negative.

Let’s define:
- $ f = -f_0 $, where $ f_0 > 0 $
- Then $ s = 3f_0 $ (since $ s = 3|f| $)

Now use thin lens equation:

$$
\frac{1}{f} = \frac{1}{s} + \frac{1}{s'}
\Rightarrow \frac{1}{-f_0} = \frac{1}{3f_0} + \frac{1}{s'}
$$

Solve for $ s' $:

$$
\frac{1}{s'} = \frac{1}{-f_0} - \frac{1}{3f_0} = -\frac{1}{f_0} - \frac{1}{3f_0} = -\frac{4}{3f_0}
\Rightarrow s' = -\frac{3f_0}{4}
$$

So:
- $ s' = -0.75f_0 $ → virtual, on the same side as object

Magnification:
$$
m = -\frac{s'}{s} = -\left( \frac{-0.75f_0}{3f_0} \right) = -(-0.25) = +0.25
$$

So:
- $ y' = m \cdot y = 0.25y $ → height is 1/4 of object
- $ m > 0 $ → upright

✔ Summary:
- $ s' = -0.75f $
- $ y' = 0.25y $
- Virtual, upright, diminished

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## ✔ Case 2: $ s = 2f $

$ s = 2f_0 $, $ f = -f_0 $

$$
\frac{1}{-f_0} = \frac{1}{2f_0} + \frac{1}{s'}
\Rightarrow \frac{1}{s'} = -\frac{1}{f_0} - \frac{1}{2f_0} = -\frac{3}{2f_0}
\Rightarrow s' = -\frac{2f_0}{3} \approx -0.667f_0
$$

Magnification:
$$
m = -\frac{s'}{s} = -\left( \frac{-2f_0/3}{2f_0} \right) = -(-1/3) = +1/3
\Rightarrow y' = \frac{1}{3}y
$$

✔ Summary:
- $ s' = -\frac{2}{3}f $
- $ y' = \frac{1}{3}y $
- Virtual, upright, diminished

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## ✔ Case 3: $ s = f $

$ s = f_0 $, $ f = -f_0 $

$$
\frac{1}{-f_0} = \frac{1}{f_0} + \frac{1}{s'}
\Rightarrow \frac{1}{s'} = -\frac{1}{f_0} - \frac{1}{f_0} = -\frac{2}{f_0}
\Rightarrow s' = -\frac{f_0}{2} = -0.5f_0
$$

Magnification:
$$
m = -\frac{s'}{s} = -\left( \frac{-0.5f_0}{f_0} \right) = -(-0.5) = +0.5
\Rightarrow y' = 0.5y
$$

✔ Summary:
- $ s' = -0.5f $
- $ y' = 0.5y $
- Virtual, upright, diminished

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## ✔ Case 4: $ s = \frac{1}{2}f $

$ s = 0.5f_0 $, $ f = -f_0 $

$$
\frac{1}{-f_0} = \frac{1}{0.5f_0} + \frac{1}{s'}
= \frac{2}{f_0} + \frac{1}{s'}
\Rightarrow \frac{1}{s'} = -\frac{1}{f_0} - \frac{2}{f_0} = -\frac{3}{f_0}
\Rightarrow s' = -\frac{f_0}{3} \approx -0.333f_0
$$

Magnification:
$$
m = -\frac{s'}{s} = -\left( \frac{-f_0/3}{0.5f_0} \right) = -(-2/3) = +2/3
\Rightarrow y' = \frac{2}{3}y
$$

Wait! That would be larger than previous cases? But still less than object.

But wait: $ m = +2/3 $ → image is smaller than object.

✔ Summary:
- $ s' = -\frac{1}{3}f $
- $ y' = \frac{2}{3}y $
- Virtual, upright, diminished

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## ✔ Case 5: $ s = \frac{1}{3}f $

$ s = \frac{1}{3}f_0 $, $ f = -f_0 $

$$
\frac{1}{-f_0} = \frac{1}{(1/3)f_0} + \frac{1}{s'} = \frac{3}{f_0} + \frac{1}{s'}
\Rightarrow \frac{1}{s'} = -\frac{1}{f_0} - \frac{3}{f_0} = -\frac{4}{f_0}
\Rightarrow s' = -\frac{f_0}{4} = -0.25f_0
$$

Magnification:
$$
m = -\frac{s'}{s} = -\left( \frac{-0.25f_0}{(1/3)f_0} \right) = -(-0.75) = +0.75
\Rightarrow y' = 0.75y
$$

✔ Summary:
- $ s' = -0.25f $
- $ y' = 0.75y $
- Virtual, upright, diminished

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❗ Important Observation:

As the object gets closer to the lens ($ s $ decreases), the image gets closer to the lens and larger, but still virtual and upright, and always smaller than object.

Even though $ y' $ increases as $ s $ decreases, it never becomes larger than object for diverging lenses.

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🧩 Ray Diagrams (Three Principal Rays):

For a diverging lens, the three principal rays are:

1. Parallel ray: A ray parallel to the principal axis → diverges as if coming from the focal point on the same side as the object.
2. Focal ray: A ray directed toward the focal point on the opposite side → emerges parallel to the axis.
3. Central ray: A ray through the center of the lens → continues straight (no bending).

⚠️ Since the image is virtual, the rays appear to diverge from a point behind the lens.

You draw these rays, extend them backward, and their intersection gives the virtual image.

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📊 Final Table Summary:

| Object Distance $ s $ | Image Distance $ s' $ | Image Height $ y' $ | Real/Virtual | Upright/Inverted |
|------------------------|-------------------------|------------------------|--------------|------------------|
| $ s = 3f $ | $ s' = -0.75f $ | $ y' = 0.25y $ | Virtual | Upright |
| $ s = 2f $ | $ s' = -0.667f $ | $ y' = 0.333y $ | Virtual | Upright |
| $ s = f $ | $ s' = -0.5f $ | $ y' = 0.5y $ | Virtual | Upright |
| $ s = 0.5f $ | $ s' = -0.333f $ | $ y' = 0.667y $ | Virtual | Upright |
| $ s = 0.333f $ | $ s' = -0.25f $ | $ y' = 0.75y $ | Virtual | Upright |

> Note: All values are in terms of $ f $, where $ f $ is the magnitude of the focal length.

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✔ Conclusion:

For any object position in front of a diverging lens, the image is:
- Always virtual
- Always upright
- Always diminished (smaller than object)
- Located between the lens and the focal point on the same side as the object

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📝 Instructions for Drawing Ray Diagrams:

For each case:
1. Draw the lens with focal points $ f $ on both sides (but remember, for diverging lens, focal points are virtual, behind the lens).
2. Draw the object as an arrow pointing up.
3. Draw:
- Ray 1: Parallel to axis → diverges as if from focal point (on same side)
- Ray 2: Toward focal point on opposite side → emerges parallel
- Ray 3: Through center → no bend
4. Extend all rays backward (dashed lines) until they intersect → this is the virtual image.

The image will be smaller, upright, and on the same side as the object.

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Let me know if you'd like a diagram sketch or help filling in the worksheet!