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GRE Subject Test: Physics : Optics

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Example Questions

Example Question #1 : Optics

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 I_o , and has an intensity of $\frac{3}{8}I_o$ after passing through both polarizes. Find $\theta$ ?

Possible Answers:

$30^\circ$

$60^\circ$

$180^\circ$

$45^\circ$

$90^\circ$

Correct answer:

$30^\circ$

Explanation:

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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Example Question #1 : Optics

A beam of light travels through a medium with index of refraction n_1=2.4 until it reaches an interface with another material, with index of refraction n_2=1.2 . 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?

Possible Answers:

$45^\circ$

This situation is impossible.

$30^\circ$

$90^\circ$

$60^\circ$

Correct answer:

$60^\circ$

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 90-30 , which is $60^\circ$ .

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Example Question #1 : Electromagnetics, Waves, And Optics

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?

Possible Answers:

$17\textup{ cm}$ to the left of the lens

No image is created

$7\textup{ cm}$ to the left of the lens

$7\textup{ cm}$ to the right of the lens

$17\textup{ cm}$ to the right of the lens

Correct answer:

$17\textup{ cm}$ to the right of the lens

Explanation:

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 d_o 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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Example Question #1 : Optics

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?

Possible Answers:

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

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

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

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

No image is created.

Correct answer:

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

Explanation:

First, find the image distance d_i 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 h_i 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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Example Question #2 : Optics

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 $600\textup{ nm}$ ?

Possible Answers:

$50\textup{ m}$

$5\textup{ mm}$

$5\textup{ cm}$

$50\textup{ cm}$

$5\textup{ m}$

Correct answer:

$5\textup{ cm}$

Explanation:

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 $5\textup{ cm}$ :

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

$D=1.22\frac{600*10^{-9}}{1.45*10^{-5}}$

$D=0.05\textup{ m}$

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Example Question #1 : Mirrors

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?

Possible Answers:

$-25 \textup{ cm}$

$-50 \textup{ cm}$

$50 \textup{ cm}$

$25 \textup{ cm}$

$2 \textup{ m}$

Correct answer:

$-25 \textup{ cm}$

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.

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GRE Subject Test: Physics : Optics

Study concepts, example questions & explanations for GRE Subject Test: Physics

All GRE Subject Test: Physics Resources

Example Questions

Example Question #1 : Optics

Example Question #1 : Optics

Example Question #1 : Electromagnetics, Waves, And Optics

Example Question #1 : Optics

Example Question #2 : Optics

Example Question #1 : Mirrors

All GRE Subject Test: Physics Resources

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