Problem Statement
Solve the optics problem: Solve the optics problem: Find the force exerted by $P = 1.0$ kW beam on a perfectly reflecting mirror. Each reflected photon transfers $2p = 2h\nu/c$ of momentum. Total force: $$F = 2P/c = 2\times1000/3\times10^8 = 6.67\times10^{-6} \text{ N} = 6.67 \mu\text{N}$$ Answer: $F = 2P/c \approx 6.7$ μN R
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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