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

Possible Answers:

\(\displaystyle 90^\circ\)

\(\displaystyle 45^\circ\)

\(\displaystyle 30^\circ\)

\(\displaystyle 180^\circ\)

\(\displaystyle 60^\circ\)

Correct answer:

\(\displaystyle 30^\circ\)

Explanation:

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

\(\displaystyle I=\frac{I_o}{2}\)

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:

\(\displaystyle I=I_o\cos^2\theta\)

Combining the two equations, we get:

\(\displaystyle I=\frac{I_o}{2}\cos^2\theta\)

Solving for \(\displaystyle \theta\):

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

Example Question #2 : Optics

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

\(\displaystyle 60^\circ\)

\(\displaystyle 45^\circ\)

\(\displaystyle 90^\circ\)

\(\displaystyle 30^\circ\)

This situation is impossible.

Correct answer:

\(\displaystyle 60^\circ\)

Explanation:

Total internal reflection occurs at the angle:

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

However, this angle is measured from a line normal to the plane of the interface; the angle we want, therefore, is \(\displaystyle 90-30\), which is \(\displaystyle 60^\circ\).

Example Question #3 : Optics

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

Possible Answers:

No image is created

\(\displaystyle 7\textup{ cm}\) to the left of the lens

\(\displaystyle 17\textup{ cm}\) to the right of the lens

\(\displaystyle 17\textup{ cm}\) to the left of the lens

\(\displaystyle 7\textup{ cm}\) to the right of the lens

Correct answer:

\(\displaystyle 17\textup{ cm}\) 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 \(\displaystyle 10-20\textup{ cm}\) on the right side of the lens.

Alternatively, one can apply the thin lens equation:

\(\displaystyle \frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}\)

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

\(\displaystyle \frac{1}{10}=\frac{1}{25} +\frac{1}{d_i}\)

\(\displaystyle \frac{1}{d_i}=\frac{5}{50}-\frac{2}{50}=\frac{3}{50}\)

\(\displaystyle d_i=\frac{50}{3}\approx 17\textup{ cm}\)

Example Question #4 : Optics

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

Possible Answers:

\(\displaystyle \frac{85}{17} \textup{ cm}\), inverted

No image is created.

\(\displaystyle \frac{17}{85} \textup{ cm}\), inverted

\(\displaystyle \frac{85}{17} \textup{ cm}\), upright

\(\displaystyle \frac{17}{85} \textup{ cm}\), upright

Correct answer:

\(\displaystyle \frac{85}{17} \textup{ cm}\), inverted

Explanation:

First, find the image distance \(\displaystyle d_i\) from the thin lens equation:

\(\displaystyle \frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}\)

\(\displaystyle \frac{1}{10}=\frac{1}{25}+\frac{1}{d_i}\)

\(\displaystyle \frac{5}{50}-\frac{2}{50}=\frac{1}{d_i}\)

\(\displaystyle d_i\approx 17\textup{ cm}\)

Magnification of a lens is given by:

\(\displaystyle M=-\frac{d_i}{d_o}=\frac{h_i}{h_o}\)

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

\(\displaystyle h_i=-\frac{d_ih_o}{d_o}=-\frac{85}{17}\textup{ cm}\)

Because the sign is negative, the image is inverted.

Example Question #5 : 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 \(\displaystyle 600\textup{ nm}\)?

Possible Answers:

\(\displaystyle 5\textup{ cm}\)

\(\displaystyle 5\textup{ mm}\)

\(\displaystyle 5\textup{ m}\)

\(\displaystyle 50\textup{ cm}\)

\(\displaystyle 50\textup{ m}\)

Correct answer:

\(\displaystyle 5\textup{ cm}\)

Explanation:

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

\(\displaystyle \theta=1.22\frac{\lambda}{D}\)

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

\(\displaystyle D=1.22\frac{\lambda}{\theta}\)

\(\displaystyle D=1.22\frac{600*10^{-9}}{1.45*10^{-5}}\)

\(\displaystyle D=0.05\textup{ m}\)

Example Question #6 : Optics

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

Possible Answers:

\(\displaystyle 2 \textup{ m}\)

\(\displaystyle -50 \textup{ cm}\)

\(\displaystyle 25 \textup{ cm}\)

\(\displaystyle -25 \textup{ cm}\)

\(\displaystyle 50 \textup{ cm}\)

Correct answer:

\(\displaystyle -25 \textup{ cm}\)

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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