Electromagnetics, Waves, and Optics

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GRE Subject Test: Physics › Electromagnetics, Waves, and Optics

Questions 1 - 10

What is the total resistance of a circuit containing four resistors ( $5\Omega, 10\Omega, 15\Omega, 20\Omega$ ) hooked up in series.

$50\Omega$

$20\Omega$

$\frac{1}{20}\Omega$

$\frac{1}{50}\Omega$

$\frac{12}{5}\Omega$

Explanation

For a circuit in series, the total resistance is simply given by the sum of each individual resistor:

$R_{Total}=R_1+R_2+R_3+...R_n$

$R_{Total}=5\Omega+10\Omega+15\Omega+20\Omega=50\Omega.$

A beam of light travels through a medium with index of refraction until it reaches an interface with another material, with index of refraction . No light is transmitted into the second material. At what angle (measured from the plane of the interface between the two materials) does the beam hit the second material?

$60^\circ$

$30^\circ$

$90^\circ$

$45^\circ$

This situation is impossible.

Explanation

Total internal reflection occurs at the angle:

$\theta=\sin^{-1}(\frac{n_2}{n_1})=\sin^{-1}(\frac{1.2}{2.4})=\sin^{-1}(\frac{1}{2})=30^o$

However, this angle is measured from a line normal to the plane of the interface; the angle we want, therefore, is , which is $60^\circ$ .

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

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 candle $5\textup{ cm}$ tall is placed $25\textup{ cm}$ to the left of a thin convex lens with focal length $10\textup{ cm}$ . What is the height and orientation of the image created?

$\frac{85}{17} \textup{ cm}$ , inverted

$\frac{85}{17} \textup{ cm}$ , upright

$\frac{17}{85} \textup{ cm}$ , upright

$\frac{17}{85} \textup{ cm}$ , inverted

No image is created.

Explanation

First, find the image distance from the thin lens equation:

$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$

$\frac{1}{10}=\frac{1}{25}+\frac{1}{d_i}$

$\frac{5}{50}-\frac{2}{50}=\frac{1}{d_i}$

$d_i\approx 17\textup{ cm}$

Magnification of a lens is given by:

$M=-\frac{d_i}{d_o}=\frac{h_i}{h_o}$

Where and are the image height and object height, respectively. The given object height is $5\textup{ cm}$ , which we can use to solve for the image height:

$h_i=-\frac{d_ih_o}{d_o}=-\frac{85}{17}\textup{ cm}$

Because the sign is negative, the image is inverted.

What is the total resistance of a circuit containing four resistors ( $5\Omega, 10\Omega, 15\Omega, 20\Omega$ ) hooked up in series.

$50\Omega$

$20\Omega$

$\frac{1}{20}\Omega$

$\frac{1}{50}\Omega$

$\frac{12}{5}\Omega$

Explanation

For a circuit in series, the total resistance is simply given by the sum of each individual resistor:

$R_{Total}=R_1+R_2+R_3+...R_n$

$R_{Total}=5\Omega+10\Omega+15\Omega+20\Omega=50\Omega.$

A reflective sphere has a diameter of $1\textup{ m}$ . The surface of the sphere makes a convex spherical mirror; what is its focal point?

$-25 \textup{ cm}$

$25 \textup{ cm}$

$50 \textup{ cm}$

$-50 \textup{ cm}$

$2 \textup{ m}$

Explanation

The focal length of a spherical mirror is one half of the radius, which is one quarter of the diameter. In the case of convex mirrors, the focal point is considered behind the surface, which gives the answer its negative sign.

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:

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

A candle $5\textup{ cm}$ tall is placed $25\textup{ cm}$ to the left of a thin convex lens with focal length $10\textup{ cm}$ . What is the height and orientation of the image created?

$\frac{85}{17} \textup{ cm}$ , inverted

$\frac{85}{17} \textup{ cm}$ , upright

$\frac{17}{85} \textup{ cm}$ , upright

$\frac{17}{85} \textup{ cm}$ , inverted

No image is created.

Explanation

First, find the image distance from the thin lens equation:

$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$

$\frac{1}{10}=\frac{1}{25}+\frac{1}{d_i}$

$\frac{5}{50}-\frac{2}{50}=\frac{1}{d_i}$

$d_i\approx 17\textup{ cm}$

Magnification of a lens is given by:

$M=-\frac{d_i}{d_o}=\frac{h_i}{h_o}$

Where and are the image height and object height, respectively. The given object height is $5\textup{ cm}$ , which we can use to solve for the image height:

$h_i=-\frac{d_ih_o}{d_o}=-\frac{85}{17}\textup{ cm}$

Because the sign is negative, the image is inverted.

A reflective sphere has a diameter of $1\textup{ m}$ . The surface of the sphere makes a convex spherical mirror; what is its focal point?

$-25 \textup{ cm}$

$25 \textup{ cm}$

$50 \textup{ cm}$

$-50 \textup{ cm}$

$2 \textup{ m}$