Refraction ray diagrams demonstrating light behavior through different materials and shapes.

To solve the problem of completing the ray diagrams for refraction, we need to apply Snell's Law and understand how light behaves when it passes through different media. Let's go through each diagram step by step.

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Snell's Law

Snell's Law relates the angles of incidence and refraction to the indices of refraction of the two media:
\[
n_1 \sin \theta_1 = n_2 \sin \theta_2
\]
where:
- \( n_1 \) is the index of refraction of the initial medium,
- \( n_2 \) is the index of refraction of the final medium,
- \( \theta_1 \) is the angle of incidence,
- \( \theta_2 \) is the angle of refraction.

For simplicity, assume:
- The index of refraction of air (\( n_1 \)) is approximately 1,
- The index of refraction of glass (\( n_2 \)) is approximately 1.5.

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Diagram 1: Rectangular Glass Block

#### Given:
- Incident ray makes an angle of \( 40^\circ \) with the normal.
- Refracted ray inside the glass block makes an angle of \( 25^\circ \) with the normal.

#### To Complete:
- Draw the refracted ray as it exits the glass block.

#### Solution:
1. Incident Ray: The incident ray enters the glass block at \( 40^\circ \) to the normal.
2. Refracted Ray Inside Glass: The refracted ray inside the glass block is already given as \( 25^\circ \) to the normal.
3. Exit from Glass Block: When the ray exits the glass block, it will bend away from the normal because it is moving from a denser medium (glass) to a less dense medium (air). Use Snell's Law again to find the exit angle:
\[
n_{\text{glass}} \sin \theta_{\text{inside}} = n_{\text{air}} \sin \theta_{\text{exit}}
\]
\[
1.5 \sin 25^\circ = 1 \sin \theta_{\text{exit}}
\]
\[
\sin \theta_{\text{exit}} = 1.5 \sin 25^\circ
\]
\[
\sin \theta_{\text{exit}} = 1.5 \times 0.4226 \approx 0.6339
\]
\[
\theta_{\text{exit}} = \sin^{-1}(0.6339) \approx 39.3^\circ
\]
So, the exit angle is approximately \( 39.3^\circ \) to the normal.

#### Final Step:
- Draw the exiting ray at \( 39.3^\circ \) to the normal on the other side of the glass block.

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Diagram 2: Horizontal Glass Surface

#### Given:
- A horizontal glass surface with a ray entering perpendicular to the surface.

#### To Complete:
- Show the path of the ray as it enters and exits the glass.

#### Solution:
1. Incident Ray: The ray is perpendicular to the surface, so the angle of incidence is \( 0^\circ \).
2. Refracted Ray Inside Glass: When the angle of incidence is \( 0^\circ \), the angle of refraction is also \( 0^\circ \). The ray continues straight without bending.
3. Exit from Glass: When the ray exits the glass, it will again be perpendicular to the surface because the angle of incidence inside the glass is \( 0^\circ \).

#### Final Step:
- Draw the ray continuing straight through the glass block without any bending.

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Diagram 3: Curved Glass Surface

#### Given:
- A curved glass surface with a ray entering at \( 42^\circ \) to the normal.

#### To Complete:
- Show the path of the ray as it refracts into the glass.

#### Solution:
1. Incident Ray: The ray enters the glass at \( 42^\circ \) to the normal.
2. Refracted Ray Inside Glass: Use Snell's Law to find the angle of refraction:
\[
n_{\text{air}} \sin \theta_{\text{incident}} = n_{\text{glass}} \sin \theta_{\text{refracted}}
\]
\[
1 \sin 42^\circ = 1.5 \sin \theta_{\text{refracted}}
\]
\[
\sin \theta_{\text{refracted}} = \frac{\sin 42^\circ}{1.5}
\]
\[
\sin \theta_{\text{refracted}} = \frac{0.6691}{1.5} \approx 0.4461
\]
\[
\theta_{\text{refracted}} = \sin^{-1}(0.4461) \approx 26.5^\circ
\]
So, the refracted ray inside the glass makes an angle of approximately \( 26.5^\circ \) to the normal.

#### Final Step:
- Draw the refracted ray inside the glass at \( 26.5^\circ \) to the normal.

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Diagram 4: Triangular Prism

#### Given:
- A triangular prism with one face at \( 45^\circ \) to the horizontal.
- A ray enters perpendicular to one face.

#### To Complete:
- Show the path of the ray as it refracts into and out of the prism.

#### Solution:
1. Incident Ray: The ray enters perpendicular to one face, so the angle of incidence is \( 0^\circ \). The ray continues straight.
2. Refracted Ray Inside Prism: The ray hits the slanted face at an angle. Use geometry to find the angle of incidence at the slanted face. Then use Snell's Law to find the angle of refraction.
3. Exit from Prism: The ray exits the prism, bending away from the normal.

#### Final Step:
- Use Snell's Law and geometry to complete the path. The exact angles depend on the geometry of the prism.

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Diagram 5: Multiple Slabs

#### Given:
- Multiple parallel slabs with rays entering at various angles.

#### To Complete:
- Show the path of the rays as they pass through the slabs.

#### Solution:
1. Incident Ray: Each ray enters a slab at a specific angle.
2. Refracted Ray Inside Slab: Use Snell's Law to find the angle of refraction inside each slab.
3. Exit from Slab: Use Snell's Law again to find the exit angle as the ray leaves each slab.

#### Final Step:
- Apply Snell's Law iteratively for each interface.

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Diagram 6: Dispersion in a Prism

#### Given:
- A triangular prism with a base angle of \( 60^\circ \).
- White light enters the prism and disperses into its component colors.

#### To Complete:
- Show the dispersion of light into its spectrum.

#### Solution:
1. Incident Ray: White light enters the prism.
2. Dispersion: Different wavelengths of light have different indices of refraction in the glass. Red light bends the least, and violet light bends the most.
3. Exit from Prism: The dispersed colors exit the prism at different angles.

#### Final Step:
- Draw the spectrum of colors (red to violet) exiting the prism.

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Final Answer

\[
\boxed{\text{Complete each diagram using Snell's Law and the principles of refraction as explained above.}}
\]