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GRE Subject Test: Physics : Lenses

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Example Questions

Example Question #1 : Optics

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

Possible Answers:

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

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

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

No image is created

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

Correct answer:

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

Alternatively, one can apply the thin lens equation:

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

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

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

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

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

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Example Question #2 : Optics

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

Possible Answers:

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

No image is created.

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

$d_i\approx 17\textup{ cm}$ $\displaystyle d_i\approx 17\textup{ cm}$

Magnification of a lens is given by:

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

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

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

Because the sign is negative, the image is inverted.

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Example Question #3 : 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 $600\textup{ nm}$ $\displaystyle 600\textup{ nm}$?

Possible Answers:

$5\textup{ mm}$ $\displaystyle 5\textup{ mm}$

$5\textup{ cm}$ $\displaystyle 5\textup{ cm}$

$50\textup{ m}$ $\displaystyle 50\textup{ m}$

$5\textup{ m}$ $\displaystyle 5\textup{ m}$

$50\textup{ cm}$ $\displaystyle 50\textup{ cm}$

Correct answer:

$5\textup{ cm}$ $\displaystyle 5\textup{ cm}$

Explanation:

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

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

Where theta is the angular resolution in radians, $\lambda$ $\displaystyle \lambda$ 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 $5\textup{ cm}$ $\displaystyle 5\textup{ cm}$:

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

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

$D=0.05\textup{ m}$ $\displaystyle D=0.05\textup{ m}$

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GRE Subject Test: Physics : Lenses

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

All GRE Subject Test: Physics Resources

Example Questions

Example Question #1 : Optics

Example Question #2 : Optics

Example Question #3 : Optics

All GRE Subject Test: Physics Resources

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