Curved Mirrors: Magnification, Image Location, and Refractive Index

Key Ideas from the Transcript

Discussion topic: curved mirrors and how to analyze them using both diagrams and numerical relationships.
Ray/diagram approach: You can use ray diagrams to visualize patterns for curved mirrors; this is useful for qualitative understanding.
Numerical approach: When numbers are involved, use the magnification equation to solve for the image location.
Specific magnification relation provided:
- The transcript states: \frac{h'}{h}= -\frac{1}{3}.
- It also states: -\frac{5}{50}= -\frac{s'}{s} which would imply a magnification of (-\frac{1}{3}) but (-\frac{5}{50}) actually equals (-0.1), not (-\tfrac{1}{3}). This is an inconsistency in the transcript.
Therefore, from the magnification equation, if \frac{h'}{h}= -\frac{1}{3} then the image-to-object distance ratio must satisfy:
\frac{s'}{s}=\frac{1}{3},
i.e., s' = \frac{s}{3}.
The transcript connects the magnification relation to the distances on the mirror, but the numeric example given contains a mismatch that should be corrected when solving problems.
The transcript then alludes to the speed of light and refractive index concepts, linking magnification discussion to wave/optics concepts through medium properties.

Fundamental relation for a curved mirror (or general imaging system): m=\frac{h'}{h}= -\frac{s'}{s}.
- Here, h' is the image height, h is the object height, s' is the image distance, and s is the object distance.
- Negative sign indicates inversion of the image relative to the object (depending on the chosen sign convention).
Given a magnification value of m= -\frac{1}{3}:
- From m= -\frac{s'}{s}, we get \frac{s'}{s}=\frac{1}{3}.
- Therefore, s' = \frac{s}{3} (the image distance is one third of the object distance).
In the transcript, the line "-5/50 equals -1/3" is mathematically incorrect (since (-5/50 = -0.1)). This indicates a misstatement; when solving, you should treat the given numerical magnification as -1/3 and derive the corresponding distance ratio accordingly.
Practical note: To locate the image using the magnification relation alone, you need at least one of the distances (either s or s') or additional information (e.g., a focal length or another relationship from the mirror equation). The transcript only provides the magnification relation, not the full mirror equation.

From the given magnification result:
- \frac{h'}{h}= -\frac{1}{3}
- \frac{s'}{s}= \frac{1}{3}
- Therefore, if the object distance is s, the image distance is s' = \frac{s}{3}.
Note on the inconsistent numeric example: The line "-5/50 = -1/3" does not hold; the correct simplification would be -\frac{5}{50} = -\frac{1}{10}, not -\frac{1}{3}. When solving problems, ensure the given magnification matches the ratio used to compute distances.

The transcript attempts to relate the magnification discussion to wave/light-speed concepts and refractive indices.
Key relation (commonly used in optics):
- The speed of light in vacuum: c \approx 3 \times 10^{8} \text{ m/s}.
- In a medium with index of refraction n, the speed is:
  v = \frac{c}{n}.
The transcript mentions the speed of light in a vacuum as c and tries to relate to the medium speed via the index of refraction, though it garbles the numeric expression (\"three times 78 meters per second\"). The correct standard form is as above.

Given: Index of refraction of water: n_{water} \approx 1.33.
Therefore, light speed in water is:
v{water} = \frac{c}{n{water}} = \frac{c}{1.33} \approx 0.752 \; c.
In numerical terms (using c \approx 3.0 \times 10^{8} \text{ m/s}):
v_{water} \approx 2.26 \times 10^{8} \text{ m/s}.
Significance: Light slows down in water by a factor of roughly 1.33, which is reflected in the refractive index; explains bending of light and imaging differences at interfaces.

How this ties to optics practice:
- Ray diagrams provide quick, qualitative understanding of image location and orientation for curved mirrors.
- The magnification equation provides a quantitative tool to relate object size, image size, and distances when enough data is available.
- Refractive index and speed in a medium explain how light propagation changes across media, which is essential when considering imaging through different materials or coatings in optical devices.
Real-world relevance:
- Designing curved mirrors (e.g., telescopes, headlights, satellite dishes) relies on understanding magnification, image location, and sign conventions.
- Understanding light speed in media is fundamental for fiber optics, lenses, and any system where light crosses media boundaries.

Magnification for a curved mirror (or image system):
m=\frac{h'}{h}= -\frac{s'}{s}.
Relation between light speed, vacuum speed, and refractive index:
v = \frac{c}{n},\quad n = \frac{c}{v}.
Index of refraction for water (given):
n_{water} \approx 1.33.
Speed of light in water (derived):
v{water} = \frac{c}{n{water}} \approx \frac{3\times 10^{8}}{1.33} \approx 2.26\times 10^{8} \text{ m/s}.$n

If \frac{h'}{h} = -\frac{1}{3}, then from the magnification relation: \frac{s'}{s} = \frac{1}{3} \Rightarrow s' = \frac{s}{3}.
The numeric line "-5/50 = -1/3" is inconsistent; the correct simplification for that line would be -\frac{5}{50} = -\frac{1}{10}. When solving, align the magnification value with the appropriate distance ratio.
The transcript attempts to link the magnification discussion to light speed and refractive index, noting water has n=1.33 and hence v_{water} = c/n \approx 0.752 c) and about 2.26 \times 10^{8} \text{ m/s}.