November 15, 2019

Exam 3

Physics 208 – Introductory Physics II
Nazareth College
Department of Chemistry & Biochemistry
Robert F Szalapski, PhD – Adjunct Lecturer
www.CallMeDrRob.com
Spring 2012

Exam 3
Sound & Geometric Optics

§ Problem 1

One firecracker makes a 95\;\mathrm{dB} sound.

§ (a)

What is the sound level for three firecrackers? (15 points)

The most direct route is to triple the intensity and to then use \log(ab)=\log(a)+\log(b).

\displaystyle 95\;\mathrm{dB} \displaystyle= \displaystyle 10\,\log\frac{I}{I_{0}}
\displaystyle\beta^{\prime} \displaystyle= \displaystyle 10\,\log\frac{3I}{I_{0}}
\displaystyle= \displaystyle 10\,\log\frac{I}{I_{0}}+10\,\log(3)
\displaystyle= \displaystyle 95\;\mathrm{dB}+4.8\;\mathrm{dB}
\displaystyle= \displaystyle 100.\;\mathrm{dB}

Alternatively we can first compute I as in part (b):

\displaystyle\beta^{\prime} \displaystyle= \displaystyle 10\,\log\frac{3I}{I_{0}}
\displaystyle= \displaystyle 10\,\log\left(3\frac{3.2\times 10^{{-3}}\frac{\mathrm{W}}{\mathrm{m}^{2}}}{1.0\times 10^{{-12}}\frac{\mathrm{W}}{\mathrm{m}^{2}}}\right)
\displaystyle= \displaystyle 100.\;\mathrm{dB}

§ (b)

What is the intensity I of the one firecracker? (15 points)

\displaystyle 95\;\mathrm{dB} \displaystyle= \displaystyle 10\,\log\frac{I}{I_{0}}
\displaystyle= \displaystyle 10\,\log\frac{I}{1.0\times 10^{{-12}}\frac{\mathrm{W}}{\mathrm{m}^{2}}}
\displaystyle I \displaystyle= \displaystyle\left(1.0\times 10^{{-12}}\frac{\mathrm{W}}{\mathrm{m}^{2}}\right)10^{{9.5}}
\displaystyle= \displaystyle 3.2\times 10^{{-3}}\frac{\mathrm{W}}{\mathrm{m}^{2}}

§ Problem 2

At the very edge of a swimming pool you shine a flashlight at the water making an angle of 45^{0} with the vertical. At what distance x does the beam hit the bottom if the pool is 2.0\;\mathrm{m} deep? (20 points)

Begin with an application of Snell’s Law:

\displaystyle 1.00\;\sin 45^{0} \displaystyle= \displaystyle 1.33\;\sin\theta
\displaystyle\theta \displaystyle= \displaystyle\sin^{{-1}}\left(\frac{1}{1.33\,\sqrt{2}}\right)
\displaystyle= \displaystyle 3\underline{2}.1^{0}

From the figure

\displaystyle\frac{x}{2.0\;\mathrm{m}} \displaystyle= \displaystyle\tan 32.1^{0}
\displaystyle x \displaystyle= \displaystyle 1.3\;\mathrm{m}

§ Problem 3

The diagram shows a large convex security mirror superimposed on a grid where each box is 10.0\;\mathrm{cm} wide.

§ (a)

If the mirror has a curvature that would be equal to a 4.00\;\mathrm{m} diameter sphere, what is the focal length, f? Plot and label f on the diagram. (5 points)

It will be a negative focal length, and its magnitude will be one-half of the radius or one-quarter of the diameter.

\displaystyle f \displaystyle= \displaystyle-\frac{1}{4}4.00\;\mathrm{m}
\displaystyle= \displaystyle-1.00\;\mathrm{m}

§ (b)

Suppose an object is placed 1.00\;\mathrm{m} in front of the mirror. Calculate the position of the image. Clearly indicate the position of the object and the image on the diagram. Is the image real or virtual? (10 points)

\displaystyle\frac{1}{d_{O}}+\frac{1}{d_{I}} \displaystyle= \displaystyle\frac{1}{f}
\displaystyle\frac{1}{1.00\;\mathrm{m}}+\frac{1}{d_{I}} \displaystyle= \displaystyle\frac{1}{-1.00\;\mathrm{m}}
\displaystyle d_{I} \displaystyle= \displaystyle-0.500\;\mathrm{m}

The image is virtual.

§ (c)

What is the magnification? (5 points)

\displaystyle M \displaystyle= \displaystyle-\frac{d_{I}}{d_{O}}
\displaystyle= \displaystyle-\frac{-0.500\;\mathrm{m}}{1.00\;\mathrm{m}}
\displaystyle= \displaystyle+0.500

§ (d)

Use ray tracing with three rays on the figure above using a straight edge to accurately show the position of the image and magnification of the object. (10 points)

§ Problem 4

The diagram shows a convex lens with a focal length f=35.0\;\mathrm{mm}.

§ (a)

Where should the object be placed if a real image is to be half the size of the object? Indicate the computed object and image on the diagram. (10 points)

The image is real, and hence it will be inverted, so the magnification is negative.

\displaystyle M \displaystyle= \displaystyle-\frac{d_{I}}{d_{O}}
\displaystyle= \displaystyle-\frac{1}{2}
\displaystyle d_{I} \displaystyle= \displaystyle\frac{1}{2}d_{O}

Hence

\displaystyle\frac{1}{d_{O}}+\frac{1}{d_{I}} \displaystyle= \displaystyle\frac{1}{f}
\displaystyle\frac{1}{d_{O}}+\frac{1}{\frac{1}{2}d_{O}} \displaystyle= \displaystyle\frac{1}{f}
\displaystyle\frac{3}{d_{O}} \displaystyle= \displaystyle\frac{1}{f}
\displaystyle d_{O} \displaystyle= \displaystyle 3f
\displaystyle= \displaystyle 3(35.0\;\mathrm{mm})
\displaystyle= \displaystyle 105\;\mathrm{mm}
\displaystyle d_{I} \displaystyle= \displaystyle\frac{1}{2}d_{O}
\displaystyle= \displaystyle\frac{1}{2}(105\;\mathrm{mm})
\displaystyle= \displaystyle 52.5\;\mathrm{mm}

§ (b)

Will the image be upright or inverted? (5 points)

Inverted.

§ (c)

Use ray tracing with three rays on the diagram above using a straight edge to clearly show the position of the image and the magnification. (10 points)

 

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