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17-Phys-A7 Optics · May 2016

Question 5 of 6: Fresnel reflection/transmission at an air–water interface

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

98-Phys-A7, Optics — National Exams, May 2016. 3 hours; closed book (formula sheet supplied). Notes 1–7 on the cover page require Q1 and Q2 (mandatory) plus any two of Q3–Q6 for a complete paper; every question is solved below as a full study resource, including all four of Q3–Q6.

Reference texts. Hecht, Optics, 5th ed.; Pedrotti, Pedrotti & Pedrotti, Introduction to Optics, 3rd ed.; Griffiths, Introduction to Electrodynamics, 4th ed. (Ch. 9, EM waves in matter).

The 6 real questions (1–6, each 15 marks) are solved in full below.

Question 5: Fresnel reflection/transmission at an air–water interface (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — normal-incidence power coefficients. Take the interface at $y=0$, normal $\hat y$, medium 1 ($n_1$) at $y<0$, medium 2 ($n_2$) at $y>0$, all waves normal, $\mathbf E\parallel\hat z$ (TE/TM coincide at normal incidence). Matching tangential $E$ and tangential $H=nE/\eta_0$ at $y=0$ gives the amplitude coefficients (same derivation family as Q4b):

$$r=\frac{n_1-n_2}{n_1+n_2},\qquad t=\frac{2n_1}{n_1+n_2}.$$

Power (irradiance) $\propto n|E|^2$, so the power reflectance compares reflected to incident power in the same medium ($n_1$ cancels), while the power transmittance must correct for the change of medium and of beam cross-section is absent at normal incidence, but the ratio of impedances is not, giving:

$$\boxed{R=r^2=\left(\frac{n_1-n_2}{n_1+n_2}\right)^{2}},\qquad \boxed{T=\frac{n_2}{n_1}t^2=1-R}$$

(the last equality is energy conservation for a lossless interface, and is the quickest way to get $T$ once $R$ is known).

Part (b) — internal reflection, water→air ($n_1=1.33,\ n_2=1.00$).

Given. $n_1=1.33$ (incidence side), $n_2=1.00$; oblique Fresnel power coefficients $R_s=r_s^2$, $R_p=r_p^2$ with $r_s=\dfrac{n_1\cos\theta_i-n_2\cos\theta_t}{n_1\cos\theta_i+n_2\cos\theta_t}$, $r_p=\dfrac{n_2\cos\theta_i-n_1\cos\theta_t}{n_2\cos\theta_i+n_1\cos\theta_t}$, $\sin\theta_t=(n_1/n_2)\sin\theta_i$.

Find. $R_s(\theta_i)$, $R_p(\theta_i)$ (with $T=1-R$) and their numerical salient values.

Approach. Evaluate the boxed normal-incidence result at $\theta_i=0$; find the (internal) Brewster angle from $R_p=0$; find the critical angle from $\sin\theta_t=1$; plot over $0\le\theta_i<\theta_c$ (beyond $\theta_c$, $\theta_t$ is complex — TIR, $R_s=R_p=1$).

  1. Normal incidence. $R_0=\left(\dfrac{1.33-1.00}{1.33+1.00}\right)^2=(0.1416)^2=\boxed{0.02006}$ (so $T_0=0.9799$).
  2. Internal Brewster angle ($R_p=0$). $\theta_{B,\text{int}}=\arctan(n_2/n_1)=\arctan(1/1.33)=\boxed{36.94^\circ}$.
  3. Critical angle ($\sin\theta_t=1$, onset of TIR). $\theta_c=\arcsin(n_2/n_1)=\arcsin(1/1.33)=\boxed{48.75^\circ}$; for $\theta_i\ge\theta_c$, $R_s=R_p=1$ (all light reflected).
0163249 0.00.51.0 angle of incidence (deg) power reflectance R Internal reflection (n1=1.33 water, n2=1.00 air) R_s (TE) R_p (TM)
Internal reflection (water to air): R_p dips to zero at the internal Brewster angle (36.9°) and both curves rise to R=1 at the critical angle (48.8°).
FeatureAngle$R_s$$R_p$
Normal incidence$0^\circ$$0.0201$$0.0201$
Internal Brewster$36.94^\circ$$\approx0.055$$0$
Critical angle$48.75^\circ$$1$$1$

Part (c) — external reflection, air→water ($n_1=1.00,\ n_2=1.33$).

Given. $n_1=1.00$, $n_2=1.33$; same Fresnel formulas, now $n_1\lt n_2$ so real $\theta_t$ exists for every $\theta_i\le90^\circ$ (no TIR).

Find. $R_s(\theta_i)$, $R_p(\theta_i)$ with salient numerical values.

  1. Normal incidence. Same $R_0=0.0201$ (the formula is symmetric in swapping $n_1\leftrightarrow n_2$).
  2. External Brewster angle. $\theta_{B,\text{ext}}=\arctan(n_2/n_1)=\arctan(1.33)=\boxed{53.06^\circ}$, where $R_p=0$ exactly — this is the more familiar "Brewster's angle" quoted for looking down into water.
  3. Grazing incidence. $\theta_i\to90^\circ\Rightarrow R_s,R_p\to1$ (a grazing surface reflects essentially all light, of either polarization) — there is no critical angle here since light always enters the denser medium.
0306090 0.00.51.0 angle of incidence (deg) power reflectance R External reflection (n1=1.00 air, n2=1.33 water) R_s (TE) R_p (TM)
External reflection (air to water): R_p dips to zero at the external Brewster angle (53.1°); both curves rise smoothly to R=1 only at grazing incidence (90°), with no critical-angle jump.
FeatureAngle$R_s$$R_p$
Normal incidence$0^\circ$$0.0201$$0.0201$
External Brewster$53.06^\circ$$\approx0.085$$0$
Grazing$90^\circ$$1$$1$

Part (d) — polarizing sunglasses. Sunlight reflecting off a horizontal surface (water, wet road, snow) near the external Brewster angle ($\approx53^\circ$ for water, roughly the angle of a low sun's glare on a lake) is reflected with $R_p\approx0$ while $R_s$ remains substantial (Part c above) — i.e. the glare is very strongly $s$-polarized, with $\mathbf E$ horizontal (parallel to the reflecting surface, perpendicular to the plane of incidence which contains the vertical). Polarizing sunglasses are therefore most effective for glare reflected near a horizontal surface at close to the Brewster angle, and are built with their transmission axis vertical so they pass the (much weaker, unpolarized-fraction) $p$-component the eye needs while blocking the strongly $s$-polarized horizontal glare.