Problem Statement
Solve the optics problem: Problem 4.194 — Waves: Sonic Crystal Negative Refraction See problem statement for all given quantities. This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematicall
Given Information
- Refractive index $n$ or focal length $f$ as given
- Object distance $u$ (negative for real objects in Cartesian convention)
- Radius of curvature $R$ or lens/mirror parameters as given
Physical Concepts & Formulas
Geometric optics is governed by Snell’s Law ($n_1 \sin\theta_1 = n_2 \sin\theta_2$) at each interface and the mirror/lens formulas in the paraxial limit. The Cartesian sign convention assigns the incident direction as positive: distances measured opposite to light are negative. For mirrors: $1/v + 1/u = 2/R = 1/f$. For thin lenses: $1/v – 1/u = 1/f$. Magnification $m = -v/u$ for mirrors and $m = v/u$ for lenses. A real image has $v > 0$ for lenses; a virtual image has $v < 0$.
- $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ — mirror formula
- $\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f}$ — thin lens formula (Cartesian)
- $n_1 \sin\theta_1 = n_2 \sin\theta_2$ — Snell’s Law
- $m = -v/u$ (mirror) or $m = v/u$ (lens) — linear magnification
Step-by-Step Solution
Step 1 — Apply correct sign convention: Real object: $u < 0$. Concave mirror/converging lens: $f > 0$.
Step 2 — Use the appropriate formula:
$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}\quad\text{(mirror)} \quad\text{or}\quad \frac{1}{v} = \frac{1}{f} + \frac{1}{u}\quad\text{(lens with Cartesian)}$$
Step 3 — Solve for image distance $v$ and compute magnification.
Worked Calculation
Substituting all values with units:
Substitute given values of $u$, $f$ into the formula and solve for $v$.
Answer
$$\boxed{\frac{1}{v} – \frac{1}{u} = \frac{1}{f}}$$
Physical Interpretation
A positive $v$ means the image forms on the real side (same side as outgoing light for lenses). Magnification $|m| > 1$ means the image is enlarged; $|m| < 1$ means diminished. A negative $m$ indicates an inverted image.
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