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

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Question

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?

Answer

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 $60^\circ$ .

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Question

What is the total resistance of a circuit consisting of three resistors in a parallel configuration? The resistors have the following resistance: $R_1=10\Omega, R_2=20\Omega, R_3=30\Omega.$

Answer

The total resistance of a circuit in parallel is given by the following equation:

$\frac{1}{R_{Total}}=\frac{1}R_{1}+\frac{1}{R_2}+\frac{1}{R_3}$

Now, we just plug in the values, and solve for the total resistance by inverting!

$\frac{1}{R_{Total}}=\frac{1}{10}+\frac{1}{20}+\frac{1}{30}$

$R_{Total}=\frac{1}{\frac{11}{60}}=\frac{60}{11}\Omega$

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Question

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

Answer

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

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

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

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Question

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?

Answer

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 $60^\circ$ .

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Question

The focal length of a thin convex lens is $10\textup{ cm}$ . A candle is placed $25\textup{ cm}$ to the left of the lens. Approximately where is the image of the candle?

Answer

Because the object is beyond 2 focal lengths of the lens, the image must be between 1 and 2 focal lengths on the opposite side. Therefore, the image must be between $10-20\textup{ cm}$ on the right side of the lens.

Alternatively, one can apply the thin lens equation:

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

Where is the object distance $(25\textup{ cm})$ and is the focal length $(10\textup{ cm})$ . Plug in these values and solve.

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

$\frac{1}{d_i}=\frac{5}{50}-\frac{2}{50}=\frac{3}{50}$

$d_i=\frac{50}{3}\approx 17\textup{ cm}$

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Question

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?

Answer

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.

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Question

Sirius is a binary star system, consisting of two white dwarfs with an angular separation of 3 arcseconds. What is the approximate minimum diameter lens needed to resolve the two stars in Sirius for an observation at $600\textup{ nm}$ ?

Answer

The Rayleigh Criterion gives the diffraction limit on resolution of a particular lens at a particular wavelength:

$\theta=1.22\frac{\lambda}{D}$

Where theta is the angular resolution in radians, $\lambda$ is the wavelength of light, and is the diameter the lens in question. Solving for and converting 3 arcseconds into radians, one can approximate the diameter to be about $5\textup{ cm}$ :

$D=1.22\frac{\lambda}{\theta}$

$D=1.22\frac{60010^{-9}}{1.4510^{-5}}$

$D=0.05\textup{ m}$

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Question

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?

Answer

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

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Question

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).

Answer

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:

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

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Question

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

Answer

The beat from two combined sound waves is:

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

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

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Question

A beam of unpolarized light passes through two linear polarizers whose polarization axes are at an angle theta with each other. The light initially has an intensity , and has an intensity of $\frac{3}{8}I_o$ after passing through both polarizes. Find $\theta$ ?

Answer

Initially unpolarized light passing though a linear polarizer decreases in intensity by a factor of two:

$I=\frac{I_o}{2}$

Malus's Law gives the change in intensity of polarized light passing through a linear polarizer in terms of the change in angle of polarization:

$I=I_o\cos^2\theta$

Combining the two equations, we get:

$I=\frac{I_o}{2}\cos^2\theta$

Solving for $\theta$ :

$\theta=\cos^{-1}(\sqrt{\frac{2I}{I_o}})=\cos^{-1}(\sqrt{\frac{2\frac{3I_o}{8}}{I_o}})=\cos^{-1}(\frac{\sqrt{3}}{2}) =30^\circ$

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Question

The focal length of a thin convex lens is $10\textup{ cm}$ . A candle is placed $25\textup{ cm}$ to the left of the lens. Approximately where is the image of the candle?

Answer

Because the object is beyond 2 focal lengths of the lens, the image must be between 1 and 2 focal lengths on the opposite side. Therefore, the image must be between $10-20\textup{ cm}$ on the right side of the lens.

Alternatively, one can apply the thin lens equation:

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

Where is the object distance $(25\textup{ cm})$ and is the focal length $(10\textup{ cm})$ . Plug in these values and solve.

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

$\frac{1}{d_i}=\frac{5}{50}-\frac{2}{50}=\frac{3}{50}$

$d_i=\frac{50}{3}\approx 17\textup{ cm}$

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Question

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?

Answer

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.

Compare your answer with the correct one above

Question

Sirius is a binary star system, consisting of two white dwarfs with an angular separation of 3 arcseconds. What is the approximate minimum diameter lens needed to resolve the two stars in Sirius for an observation at $600\textup{ nm}$ ?

Answer

The Rayleigh Criterion gives the diffraction limit on resolution of a particular lens at a particular wavelength:

$\theta=1.22\frac{\lambda}{D}$

Where theta is the angular resolution in radians, $\lambda$ is the wavelength of light, and is the diameter the lens in question. Solving for and converting 3 arcseconds into radians, one can approximate the diameter to be about $5\textup{ cm}$ :

$D=1.22\frac{\lambda}{\theta}$

$D=1.22\frac{60010^{-9}}{1.4510^{-5}}$

$D=0.05\textup{ m}$

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Question

The focal length of a thin convex lens is $10\textup{ cm}$ . A candle is placed $25\textup{ cm}$ to the left of the lens. Approximately where is the image of the candle?

Answer

Because the object is beyond 2 focal lengths of the lens, the image must be between 1 and 2 focal lengths on the opposite side. Therefore, the image must be between $10-20\textup{ cm}$ on the right side of the lens.

Alternatively, one can apply the thin lens equation:

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

Where is the object distance $(25\textup{ cm})$ and is the focal length $(10\textup{ cm})$ . Plug in these values and solve.

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

$\frac{1}{d_i}=\frac{5}{50}-\frac{2}{50}=\frac{3}{50}$

$d_i=\frac{50}{3}\approx 17\textup{ cm}$

Compare your answer with the correct one above

Question

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?

Answer

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.

Compare your answer with the correct one above

Question

Sirius is a binary star system, consisting of two white dwarfs with an angular separation of 3 arcseconds. What is the approximate minimum diameter lens needed to resolve the two stars in Sirius for an observation at $600\textup{ nm}$ ?

Answer

The Rayleigh Criterion gives the diffraction limit on resolution of a particular lens at a particular wavelength:

$\theta=1.22\frac{\lambda}{D}$

Where theta is the angular resolution in radians, $\lambda$ is the wavelength of light, and is the diameter the lens in question. Solving for and converting 3 arcseconds into radians, one can approximate the diameter to be about $5\textup{ cm}$ :

$D=1.22\frac{\lambda}{\theta}$

$D=1.22\frac{60010^{-9}}{1.4510^{-5}}$

$D=0.05\textup{ m}$

Compare your answer with the correct one above

Question

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?

Answer

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 $60^\circ$ .

Compare your answer with the correct one above

Question

The focal length of a thin convex lens is $10\textup{ cm}$ . A candle is placed $25\textup{ cm}$ to the left of the lens. Approximately where is the image of the candle?

Answer

Because the object is beyond 2 focal lengths of the lens, the image must be between 1 and 2 focal lengths on the opposite side. Therefore, the image must be between $10-20\textup{ cm}$ on the right side of the lens.

Alternatively, one can apply the thin lens equation:

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

Where is the object distance $(25\textup{ cm})$ and is the focal length $(10\textup{ cm})$ . Plug in these values and solve.

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

$\frac{1}{d_i}=\frac{5}{50}-\frac{2}{50}=\frac{3}{50}$

$d_i=\frac{50}{3}\approx 17\textup{ cm}$

Compare your answer with the correct one above

Question

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?

Answer

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.

Compare your answer with the correct one above

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