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GRE Subject Test: Physics › Waves

Questions 1 - 10

A beam of light, with wavelength $400\textup{ nm}$ , is normally incident on a transmission diffraction grating. With respect to the incident beam, the first order diffraction maximum occurs at an angle of $6^\circ$ . What is the number of slits per centimeter on the grating?

$2.5*10^3\ \frac{\textup{lines}}{\textup{cm}}$

$2.5*10^5\ \frac{\textup{lines}}{\textup{cm}}$

$4*10^6\ \frac{\textup{lines}}{\textup{cm}}$

$4*10^4\ \frac{\textup{lines}}{\textup{cm}}$

$4*10^5\ \frac{\textup{lines}}{\textup{cm}}$

Explanation

The equation describing maxima of a diffraction grating is:

$d\sin(\theta)=m\theta$

Where d is the separation of slits, which can also be expressed as:

$d=\frac{L}{N}$

Substituting d into the first equation and making the small angle approximation, one can solve for $\frac{N}{L}$ (lines per length):

$\frac{N}{L}=\frac{\theta}{m\lambda}=\frac{0.1}{1\times 400\times 10^{-9}}=2.5\times 10^5\ \frac{\textup{lines}}{\textup{m}}$

Note that the angle had to be converted from degrees to radians. Finally, the question asks for lines per centimeter, not meter, so the answer becomes $2.5*10^3\ \frac{\textup{lines}}{\textup{cm}}$ .

$2.5*10^3\ \frac{\textup{lines}}{\textup{cm}}$

$2.5*10^5\ \frac{\textup{lines}}{\textup{cm}}$

$4*10^6\ \frac{\textup{lines}}{\textup{cm}}$

$4*10^4\ \frac{\textup{lines}}{\textup{cm}}$

$4*10^5\ \frac{\textup{lines}}{\textup{cm}}$

Explanation

The equation describing maxima of a diffraction grating is:

$d\sin(\theta)=m\theta$

Where d is the separation of slits, which can also be expressed as:

$d=\frac{L}{N}$

Substituting d into the first equation and making the small angle approximation, one can solve for $\frac{N}{L}$ (lines per length):

$\frac{N}{L}=\frac{\theta}{m\lambda}=\frac{0.1}{1\times 400\times 10^{-9}}=2.5\times 10^5\ \frac{\textup{lines}}{\textup{m}}$

Note that the angle had to be converted from degrees to radians. Finally, the question asks for lines per centimeter, not meter, so the answer becomes $2.5*10^3\ \frac{\textup{lines}}{\textup{cm}}$ .

A car's horn has a frequency of $439\textup{ Hz}$ . As the car drives away from you, you hear the horn at a frequency of $384\textup{ Hz}$ . What is the speed of the car?

$v_{sound}=343\ \frac{\textup{m}}{\textup{s}}$

$49.13\ \frac{\textup{m}}{\textup{s}}$

$43.87\ \frac{\textup{m}}{\textup{s}}$

$40\ \frac{\textup{m}}{\textup{s}}$

$37.45\ \frac{\textup{m}}{\textup{s}}$

$52\ \frac{\textup{m}}{\textup{s}}$

Explanation

The equation for the Doppler effect with frequency for a moving source is given by:

$f'=\frac{f}{1\pm \frac{v_o}{v_s}}$

Where is the speed of sound, is the speed of the source, is the stationary frequency and is the Doppler frequency. Substituting the known values and solving gives us the speed of the car.

$384=\frac{439}{1\pm \frac{v_o}{343}}$

$384=439*\frac{343}{343\pm v_o}$

$384=\frac{150577}{343\pm v_o}$

$131712\pm 384v_o=150577$

$v_o=49.13\ \frac{\textup{m}}{\textup{s}}$

A double slit experiment is set up with the following parameters: two slits are a separated by a distance . A beam of light with wavelength $\lambda$ shines through the two slits, and is projected onto a screen a distance from the slits. What is the distance on the screen between the central band and the next band on either side? (This distance is marked '' on the figure).

Slit

$x=\frac{L\lambda}{d}$

$x=\frac{2L\lambda}{d}$

$x=\frac{L^2\lambda}{d^2}$

$x=\frac{2L^2\lambda}{d^2}$

$x=\frac{L^2\lambda}{2d^2}$

Explanation

The condition for constructive interference with double slit diffraction is given by:

$d\sin(\theta)=m\lambda$

Where is 0, 1, 2, ...

Solving for the angle and using the small angle approximation, we get;

$\sin(\theta)=\frac{m\lambda}{d}\approx \theta$

The distance in the diagram can be related to the other quantities by simple geometry:

$\tan (\theta)=\frac{x}{L}\approx \theta$

Again, with the help of the small angle approximation. Setting the two thetas equal to each other and solving for , we get:

$x=\frac{Lm\lambda}{d}$

For the central band, , so the position is also zero. The next band, , yields a distance of:

$x=\frac{L\lambda}{d}$

Slit

$x=\frac{L\lambda}{d}$

$x=\frac{2L\lambda}{d}$

$x=\frac{L^2\lambda}{d^2}$

$x=\frac{2L^2\lambda}{d^2}$

$x=\frac{L^2\lambda}{2d^2}$

Explanation

The condition for constructive interference with double slit diffraction is given by:

$d\sin(\theta)=m\lambda$

Where is 0, 1, 2, ...

Solving for the angle and using the small angle approximation, we get;

$\sin(\theta)=\frac{m\lambda}{d}\approx \theta$

The distance in the diagram can be related to the other quantities by simple geometry:

$\tan (\theta)=\frac{x}{L}\approx \theta$

Again, with the help of the small angle approximation. Setting the two thetas equal to each other and solving for , we get:

$x=\frac{Lm\lambda}{d}$

For the central band, , so the position is also zero. The next band, , yields a distance of:

$x=\frac{L\lambda}{d}$

A car's horn has a frequency of $439\textup{ Hz}$ . As the car drives away from you, you hear the horn at a frequency of $384\textup{ Hz}$ . What is the speed of the car?

$v_{sound}=343\ \frac{\textup{m}}{\textup{s}}$

$49.13\ \frac{\textup{m}}{\textup{s}}$

$43.87\ \frac{\textup{m}}{\textup{s}}$

$40\ \frac{\textup{m}}{\textup{s}}$

$37.45\ \frac{\textup{m}}{\textup{s}}$

$52\ \frac{\textup{m}}{\textup{s}}$

Explanation

The equation for the Doppler effect with frequency for a moving source is given by:

$f'=\frac{f}{1\pm \frac{v_o}{v_s}}$

Where is the speed of sound, is the speed of the source, is the stationary frequency and is the Doppler frequency. Substituting the known values and solving gives us the speed of the car.

$384=\frac{439}{1\pm \frac{v_o}{343}}$

$384=439*\frac{343}{343\pm v_o}$

$384=\frac{150577}{343\pm v_o}$

$131712\pm 384v_o=150577$

$v_o=49.13\ \frac{\textup{m}}{\textup{s}}$

Two waves with frequencies: $f_1=6 \textup{ MHz}, \ f_2=2 \textup{ MHz}$ are combined. What is the frequency of the resulting beat?

$4\textup{ MHz}$

$8 \textup{ MHz}$

$12: \textup{MHz}$

$3 \textup{ MHz}$

$2\sqrt{10} \textup{ MHz}$

Explanation

The beat from two combined sound waves is:

$f_{beat}=|f_1-f_2|$

$f_{beat}=|6000-2000|=4\textup{ MHz}$

A monochromatic beam of $600\textup{ nm}$ light travels through a material with a phase velocity of $2*10^8\ \frac{\textup{m}}{\textup{s}}$ . What is the refractive index of the material?

Explanation

The phase velocity of light through a medium with refractive index is given by:

$v_p=\frac{c}{n}$

Solving this for and substituting our given value of :

$n=\frac{c}{v_p}=\frac{3\times10^8}{2\times10^8}=1.5$

Two waves with frequencies: $f_1=6 \textup{ MHz}, \ f_2=2 \textup{ MHz}$ are combined. What is the frequency of the resulting beat?

$4\textup{ MHz}$

$8 \textup{ MHz}$

$12: \textup{MHz}$

$3 \textup{ MHz}$

$2\sqrt{10} \textup{ MHz}$

Explanation

The beat from two combined sound waves is:

$f_{beat}=|f_1-f_2|$

$f_{beat}=|6000-2000|=4\textup{ MHz}$

A monochromatic beam of $600\textup{ nm}$ light travels through a material with a phase velocity of $2*10^8\ \frac{\textup{m}}{\textup{s}}$ . What is the refractive index of the material?

Explanation

The phase velocity of light through a medium with refractive index is given by:

$v_p=\frac{c}{n}$

Solving this for and substituting our given value of :

$n=\frac{c}{v_p}=\frac{3\times10^8}{2\times10^8}=1.5$

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