Category: Part 5: Optics

  • Problem 5.133 — Wedge Film: Angle from Fringe Count

    Problem Statement An air wedge shows 15 dark fringes per cm in reflected light for $\lambda = 600$ nm. Find the wedge angle $\alpha$. Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws…

  • Problem 5.131 — Interference: Source Size and Coherence

    Problem Statement Solve the oscillation/wave problem: An extended source of width $b = 0.5$ mm is 1.0 m from a double slit ($d = 0.5$ mm). For $\lambda = 500$ nm, determine whether the fringes are visible. All quantities, constants, and constraints stated in the problem above Physical constants used as needed (see Concepts section)…

  • Problem 5.130 — Interference Filter Transmission

    Problem Statement Solve the oscillation/wave problem: An interference filter consists of a thin film of MgF$_2$ ($n = 1.38$) between two silver mirrors. Find the thickness for maximum transmission at $\lambda = 550$ nm. The filter transmits when $2nt = m\lambda$ (resonance condition, both reflections cause phase shifts of $\pi$ which cancel). Minimum t Given…

  • Problem 5.129 — Optical Rotation: Quartz

    Problem Statement Solve the optics problem: A quartz crystal of thickness $t = 1.0$ mm rotates the plane of polarization by $\phi = 21.7°$ at $\lambda = 589$ nm. Calculate the specific rotation $[\alpha]$. $$[\alpha] = \frac{\phi}{t} = \frac{21.7°}{1.0\text{ mm}} = \boxed{21.7°/\text{mm} = 217°/\text{cm}}$$ Given Information $[\alpha] = \frac{\phi}{t} = \frac{21.7°}{1.0\text{ mm}$ Physical Concepts &…

  • Problem 5.127 — Absorption Coefficient from Transmittance

    Problem Statement A glass plate of thickness $t = 3.0$ mm transmits $T = 72\%$ of normally incident light (ignoring reflection losses). Find the absorption coefficient. Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant…

  • Problem 5.128 — Scattering: Angular Distribution

    Problem Statement Rayleigh scattering of sunlight. Find the ratio of scattered intensity at $\theta = 90°$ to that at $\theta = 0°$ for a single dipole oscillator. Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics principles to the scenario described. The solution requires identifying the…

  • Problem 5.126 — Photometry: Contrast Ratio

    Problem Statement A cinema screen has luminance $L_w = 100$ cd/m² for a white area and $L_b = 0.5$ cd/m² for a black area. Find the contrast ratio. Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics principles to the scenario described. The solution requires identifying…

  • Problem 5.124 — Double Refraction: e-ray and o-ray Separation

    Problem Statement Solve the optics problem: A calcite crystal of thickness $t = 5.0$ mm ($n_o = 1.658$, $n_e = 1.486$) is cut with the optic axis at $45°$ to the surface. Find the lateral separation between the ordinary and extraordinary rays at normal incidence. At normal incidence to a calcite plate cut with optic…

  • Problem 5.125 — Babinet’s Principle

    Problem Statement A circular disk of diameter $2a = 1.0$ mm blocks the centre of a beam. The observation point is on the axis at distance $b = 1.0$ m. Wavelength $\lambda = 500$ nm. What is the intensity at the centre of the geometric shadow? Given Information See problem statement for all given quantities.…

  • Problem 5.123 — Refractive Index Measurement by Minimum Deviation

    Problem Statement Solve the optics problem: A glass prism (apex angle $\Theta = 45°$) gives minimum deviation $\delta_m = 30°$. Find its refractive index. $$n = \frac{\sin\frac{\Theta+\delta_m}{2}}{\sin\frac{\Theta}{2}} = \frac{\sin\frac{75°}{2}}{\sin\frac{45°}{2}} = \frac{\sin 37.5°}{\sin 22.5°} = \frac{0.6088}{0.3827} \approx \boxed{1.59}$$ Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics…