A ray of white light in air is incident normally ($\theta_1 = 0^\circ$) on the first surface of a glass prism with apex angle $\Phi$, where $\Phi$ is less than the critical angle for the glass-air interface. The glass has an index of refraction $n$ that depends on wavelength. The ray travels through the glass and refracts as it emerges from the second surface into air, deviating from its original path by a total angle $\delta$. | Choice | Total Angle of Deviation $\delta$ | Wavelength with Greater Deviation | | :---: | :---: | :---: | | A | $\arcsin(n \sin\Phi) + \Phi$ | Violet light | | B | $\arcsin(n \sin\Phi) - \Phi$ | Violet light | | C | $\arcsin\left(\dfrac{\sin\Phi}{n}\right) - \Phi$ | Red light | | D | $\arcsin(n \sin\Phi) - \Phi$ | Red light | Which entry in the table correctly gives the expression for the total angle of deviation $\delta$ and identifies which visible component of the light experiences the greater deviation? • Nerd Notes

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Advanced

Step	Reasoning
Determine the total angle of deviation $\delta$ and compare deviation between red and violet light using Snell’s law and geometry. \[ n_1 \sin\theta_1 = n_2 \sin\theta_2 \]	The question asks for a symbolic expression for $\delta$ and a qualitative comparison of wavelength deviation.
Determine the angle of incidence at the second surface inside the prism. \[ \theta_{\text{inc}} = 90^\circ – (90^\circ – \Phi) = \Phi \]	Since the ray strikes the first face normally ($\theta_1 = 0^\circ$), it passes through without bending. In the right triangle formed by the top apex $\Phi$, the first face, and the ray path, the angle between the ray and the second face is $90^\circ – \Phi$. Therefore, the angle of incidence $\theta_{\text{inc}}$ relative to the normal of the second surface is $\Phi$.
Apply Snell’s law at the glass-air interface at the second surface to find the exit angle $\theta_2$. \[ n \sin\Phi = 1 \cdot \sin\theta_2 \implies \theta_2 = \arcsin(n \sin\Phi) \]	Light travels from glass (index $n$) to air (index $1$).
Calculate the total deviation angle $\delta$. \[ \delta = \theta_2 – \Phi = \arcsin(n \sin\Phi) – \Phi \]	The ray was traveling at angle $\Phi$ relative to the surface normal, and emerges at angle $\theta_2$ relative to the normal. The angular change from its straight-line path is the difference between these angles.
Compare deviation for red vs. violet light using optical dispersion. \[ n_{\text{violet}} > n_{\text{red}} \implies \arcsin(n_{\text{violet}}\sin\Phi) – \Phi > \arcsin(n_{\text{red}}\sin\Phi) – \Phi \implies \delta_{\text{violet}} > \delta_{\text{red}} \]	In glass, shorter wavelengths (violet light, $\lambda \approx 400\text{ nm}$) interact more strongly with the medium than longer wavelengths (red light, $\lambda \approx 700\text{ nm}$), giving glass a higher refractive index for violet light ($n_{\text{violet}} > n_{\text{red}}$).

Why each choice is correct or incorrect:

(A) Incorrectly adds the incident angle $\Phi$ to the exit angle $\theta_2$ instead of subtracting it to find the angle of deviation.

(B) This is the correct answer, properly applying Snell’s law at the second interface and accounting for optical dispersion where shorter wavelengths experience a higher refractive index.

(C) Inverts Snell’s law by dividing by $n$ when transitioning from glass to air and misidentifies red light as having a higher refractive index than violet light.

(D) Correctly derives the expression for $\delta$ but incorrectly assumes red light has a higher index of refraction than violet light, leading to an incorrect prediction about dispersion.

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Step	Reasoning
Determine the total angle of deviation \(\delta\) and compare deviation between red and violet light using Snell’s law and geometry. \[ n_1 \sin\theta_1 = n_2 \sin\theta_2 \]	The question asks for a symbolic expression for \(\delta\) and a qualitative comparison of wavelength deviation.
Determine the angle of incidence at the second surface inside the prism. \[ \theta_{\text{inc}} = 90^\circ – (90^\circ – \Phi) = \Phi \]	Since the ray strikes the first face normally (\(\theta_1 = 0^\circ\)), it passes through without bending. In the right triangle formed by the top apex \(\Phi\), the first face, and the ray path, the angle between the ray and the second face is \(90^\circ – \Phi\). Therefore, the angle of incidence \(\theta_{\text{inc}}\) relative to the normal of the second surface is \(\Phi\).
Apply Snell’s law at the glass-air interface at the second surface to find the exit angle \(\theta_2\). \[ n \sin\Phi = 1 \cdot \sin\theta_2 \implies \theta_2 = \arcsin(n \sin\Phi) \]	Light travels from glass (index \(n\)) to air (index \(1\)).
Calculate the total deviation angle \(\delta\). \[ \delta = \theta_2 – \Phi = \arcsin(n \sin\Phi) – \Phi \]	The ray was traveling at angle \(\Phi\) relative to the surface normal, and emerges at angle \(\theta_2\) relative to the normal. The angular change from its straight-line path is the difference between these angles.
Compare deviation for red vs. violet light using optical dispersion. \[ n_{\text{violet}} > n_{\text{red}} \implies \arcsin(n_{\text{violet}}\sin\Phi) – \Phi > \arcsin(n_{\text{red}}\sin\Phi) – \Phi \implies \delta_{\text{violet}} > \delta_{\text{red}} \]	In glass, shorter wavelengths (violet light, \(\lambda \approx 400\text{ nm}\)) interact more strongly with the medium than longer wavelengths (red light, \(\lambda \approx 700\text{ nm}\)), giving glass a higher refractive index for violet light (\(n_{\text{violet}} > n_{\text{red}}\)).

Kinematics	Forces
\(\Delta x = v_i t + \frac{1}{2} at^2\)	\(F = ma\)
\(v = v_i + at\)	\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)	\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)	\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\)

Circular Motion	Energy
\(F_c = \frac{mv^2}{r}\)	\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)	\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)	\(KE_i + PE_i = KE_f + PE_f\)
	\(W = Fd \cos\theta\)

Momentum	Torque and Rotations
\(p = mv\)	\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)	\(I = \sum mr^2\)
\(p_i = p_f\)	\(L = I \cdot \omega\)

Simple Harmonic Motion	Fluids
\(F = -kx\)	\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)	\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)	\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)	\(F_b = \rho V g\)
\(a = -\omega^2 x\)	\(A_1v_1 = A_2v_2\)

Constant	Description
[katex]g[/katex]	Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]	Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]	Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]	Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]	Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]	Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]	Mass of the Sun

Variable	SI Unit
[katex]s[/katex] (Displacement)	[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)	[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)	[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)	[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)	[katex]\text{kilograms (kg)}[/katex]

Variable	Derived SI Unit
[katex]F[/katex] (Force)	[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)	[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)	[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)	[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)	[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)	[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)	[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)	[katex]\text{hertz (Hz)}[/katex]

Prefix	Symbol	Power of Ten
Pico-	p	[katex]10^{-12}[/katex]
Nano-	n	[katex]10^{-9}[/katex]
Micro-	µ	[katex]10^{-6}[/katex]
Milli-	m	[katex]10^{-3}[/katex]
Centi-	c	[katex]10^{-2}[/katex]
Deci-	d	[katex]10^{-1}[/katex]
(Base unit)	–	[katex]10^{0}[/katex]
Deca- or Deka-	da	[katex]10^{1}[/katex]
Hecto-	h	[katex]10^{2}[/katex]
Kilo-	k	[katex]10^{3}[/katex]
Mega-	M	[katex]10^{6}[/katex]
Giga-	G	[katex]10^{9}[/katex]
Tera-	T	[katex]10^{12}[/katex]

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AP Physics

Unit 1 - Vectors and Kinematics

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Advanced

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