Superposition and Interference — Hard Practice Quiz

A Physics cheat sheet for Superposition and Interference — every key formula with its symbols defined — plus a hard-level practice quiz to test recall.

Practice quiz

1. Light of wavelength $\lambda$ passes through a single slit of width $D$, producing its first diffraction minimum at an angle $\theta$. If a sound wave with this same wavelength $\lambda$ is used to excite an open-open tube of length $L$, what is the ratio of the tube's fundamental frequency to the speed of sound $v$ in terms of $D$ and $\theta$?

   • $\frac{1}{D \sin \theta}$
   • $\frac{2L}{D \sin \theta}$
   • $\frac{D \sin \theta}{2L}$
   • $\frac{1}{2L D \sin \theta}$

   Answer: $\frac{1}{D \sin \theta}$

2. A circular aperture of diameter $D_A$ produces its first diffraction minimum at an angle $\theta$ for light of wavelength $\lambda_L$. A sound wave of wavelength $\lambda_S$ is used to excite an open-closed tube of length $L_T$. If $\lambda_L = \lambda_S$ and the fundamental frequency of the tube is $f_1$, which of the following expressions correctly represents the diameter $D_A$ in terms of $L_T$, $v$ (speed of sound), and $\theta$?

   • $\frac{1.22 v}{4 L_T \sin \theta}$
   • $\frac{4.88 L_T}{\sin \theta}$
   • $\frac{1.22 L_T \sin \theta}{v}$
   • $\frac{4 L_T \sin \theta}{1.22}$

   Answer: $\frac{4.88 L_T}{\sin \theta}$

3. An open-open tube of length $L_1$ has a fundamental frequency $f_{oo}$. An open-closed tube of length $L_2$ has a fundamental frequency $f_{oc}$. If the wavelength of the sound producing $f_{oo}$ is equal to the wavelength of light that produces the first minimum for a single slit of width $D$ at angle $\theta$, and the wavelength of the sound producing $f_{oc}$ is twice that, what is the ratio $\frac{L_1}{L_2}$ if $f_{oo} = f_{oc}$?

   • $1/2$
   • $1$
   • $2$
   • $4$

   Answer: $2$

4. Light of wavelength $\lambda_1$ passes through a single slit of width $D$, producing its first diffraction minimum at an angle $\theta$. This wavelength $\lambda_1$ is also the fundamental wavelength of sound in an open-open tube of length $L$. If the slit width is doubled to $2D$, and a new light wavelength $\lambda_2$ is used such that it produces the *same* first minimum angle $\theta$, how does the new fundamental frequency $f_2$ of the open-open tube (excited by $\lambda_2$) compare to the original fundamental frequency $f_1$?

   • $f_2 = 2 f_1$
   • $f_2 = f_1$
   • $f_2 = \frac{1}{2} f_1$
   • $f_2 = \frac{1}{4} f_1$

   Answer: $f_2 = \frac{1}{2} f_1$

5. An open-closed tube of length $L$ is resonating at its third harmonic. The sound wavelength produced by this harmonic is $\lambda_S$. If light of this same wavelength $\lambda_S$ passes through a circular aperture of diameter $D$, which of the following expressions correctly represents $\sin \theta$, where $\theta$ is the angle to the first diffraction minimum?

   • $\frac{1.22 L}{D}$
   • $\frac{1.22 \cdot 4L}{3D}$
   • $\frac{1.22 \cdot 3L}{4D}$
   • $\frac{1.22 L}{3D}$

   Answer: $\frac{1.22 \cdot 4L}{3D}$

6. Light of wavelength $\lambda_1$ passes through a single slit of width $D_1$, producing a first minimum at angle $\theta$. This wavelength $\lambda_1$ corresponds to the fundamental wavelength of sound in an open-open tube of length $L_1$. Light of wavelength $\lambda_2$ passes through a single slit of width $D_2$, producing the *same* first minimum angle $\theta$. This wavelength $\lambda_2$ corresponds to the third harmonic of an open-closed tube of length $L_2$. If $L_1 = L_2 = L$, what is the ratio $\frac{D_1}{D_2}$?

   • $2/3$
   • $3/2$
   • $1/2$
   • $4/3$

   Answer: $3/2$

7. An open-open tube of length $L$ produces its fundamental frequency $f_1$. The wavelength of this sound is used as the wavelength of light for a single slit diffraction experiment, producing a first minimum at angle $\theta$. If the length of the tube is doubled to $2L$, how does the new first minimum angle $\theta'$ compare to the original angle $\theta$, assuming the slit width $D$ remains constant?

   • $\sin \theta' = \frac{1}{2} \sin \theta$
   • $\sin \theta' = \sin \theta$
   • $\sin \theta' = 2 \sin \theta$
   • $\sin \theta' = 4 \sin \theta$

   Answer: $\sin \theta' = 2 \sin \theta$

8. Light of wavelength $\lambda_L$ passes through a single slit of width $D = 2 \times 10^{-5} \text{ m}$, producing a first minimum at an angle $\theta$ such that $\sin \theta = 0.03$. This wavelength $\lambda_L$ is equal to the wavelength of the second harmonic of an open-open tube. If the speed of sound in the tube is $v = 340 \text{ m/s}$, what is the length $L$ of the tube?

   • $6 \times 10^{-7} \text{ m}$
   • $1.2 \times 10^{-6} \text{ m}$
   • $3 \times 10^{-7} \text{ m}$
   • $6 \times 10^{-5} \text{ m}$

   Answer: $6 \times 10^{-7} \text{ m}$

9. An open-open tube of length $L$ produces its fundamental frequency. The wavelength of this sound is $\lambda$. This wavelength $\lambda$ is then used for two diffraction experiments: first, through a single slit of width $D_S$, producing a first minimum at angle $\theta_S$; second, through a circular aperture of diameter $D_C$, producing a first minimum at angle $\theta_C$. If the slit width $D_S$ is equal to the circular aperture diameter $D_C$, what is the ratio $\frac{\sin \theta_C}{\sin \theta_S}$?

   • $1$
   • $1.22$
   • $1/1.22$
   • $2$

   Answer: $1.22$

10. An open-open tube of length $L_{oo}$ produces its fundamental frequency. The wavelength corresponding to this frequency is used for a single slit diffraction experiment with slit width $D_S$, producing a first minimum at angle $\theta_S$. An open-closed tube of length $L_{oc}$ produces its third harmonic frequency. The wavelength corresponding to this frequency is used for a circular aperture diffraction experiment with diameter $D_C$, producing a first minimum at angle $\theta_C$. If $L_{oo} = L_{oc} = L$, and $\sin \theta_S = \sin \theta_C$, what is the ratio $\frac{D_S}{D_C}$?

    • $\frac{1.22 \cdot 4}{6}$
    • $\frac{6}{1.22 \cdot 4}$
    • $\frac{1.22 \cdot 3}{2}$
    • $\frac{2}{1.22 \cdot 3}$

    Answer: $\frac{6}{1.22 \cdot 4}$