November 15, 2019

Solution Set 6

Physics 208-02 – Introductory Physics II
Solution Set 6
Nazareth College
Department of Chemistry & Biochemistry
Robert F Szalapski, PhD – Adjunct Lecturer
www.CallMeDrRob.com
Spring 2012

§ Chapter 23, Problem 4

A person whose eyes are 1.68\;\mathrm{m} above the floor stands 2.30\;\mathrm{m} in front of a vertical plane mirror whose bottom edge is 43\;\mathrm{cm} above the floor. What is the horizontal distance x between the wall and the point on the floor nearest to the wall that can be seen reflected in the mirror?

Notice that incident and reflected angles are equal, hence they have been marked as \theta. Their complementary angles, which will also be equal, have been marked as \phi. The triangle marked in red is then similar to the triangle marked in green by AAA similarity. This means that ratios of corresponding sides are equal. Hence

\displaystyle\frac{x}{43\;\mathrm{cm}} \displaystyle= \displaystyle\frac{230\;\mathrm{cm}}{125\;\mathrm{cm}}
\displaystyle x \displaystyle= \displaystyle 79\;\mathrm{cm}

Virtually all of the work is in the construction of the figure.

§ Chapter 23, Problem 9

If you look at yourself in a shiny Christmas-tree ball with a diameter of 9.0\;\mathrm{cm} when your face is 30.0\;\mathrm{cm} from it, where is your image? Is it real or virtual? Is it upright or inverted?

The figure has been drawn roughly to scale except for the size of the cartoon head. Incident and reflected rays have been paired by colors. Notice that the focal length is one-half of the radius which in turn is is one-half of the diameter, so the focal length is one-quarter of the diameter. It is also negative because this is a convex mirror.

\displaystyle\frac{1}{d_{O}}+\frac{1}{d_{I}} \displaystyle= \displaystyle\frac{1}{f}
\displaystyle\frac{1}{30.0\;\mathrm{cm}}+\frac{1}{d_{I}} \displaystyle= \displaystyle\frac{1}{-2.\underline{2}5\;\mathrm{cm}}
\displaystyle d_{I} \displaystyle= \displaystyle-2.1\;\mathrm{cm}

The image appears to be inside of the ball by 2.1\;\mathrm{cm}, so the image is virtual. The magnification is given by

\displaystyle M \displaystyle= \displaystyle-\frac{d_{I}}{d_{O}}
\displaystyle= \displaystyle-\frac{-2.1\;\mathrm{cm}}{30.0\;\mathrm{cm}}
\displaystyle= \displaystyle+0.070

The image is upright.

§ Chapter 23, Problem 14

You are standing 3.0\;\mathrm{m} in front of a security mirror in a store. You estimate the height of your image to be one-half of your actual height. Estimate the radius of curvature of the mirror.

First look at the magnification equation.

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

Now we can find the focal length as

\displaystyle\frac{1}{f} \displaystyle= \displaystyle\frac{1}{d_{O}}+\frac{1}{d_{I}}
\displaystyle= \displaystyle\frac{1}{3.0\;\mathrm{m}}+\frac{1}{-1.5\mathrm{m}}
\displaystyle f= \displaystyle-3.0\;\mathrm{m}
\displaystyle r \displaystyle= \displaystyle 6.0\;\mathrm{m}

§ Chapter 23, Problem 20

The magnification of a convex mirror is +0.65x for objects 2.2\;\mathrm{m} from the mirror. What is the focal length of the mirror?

Any diagram would be very similar to the one in Problem 14. Let’s use the magnification to find the image position.

\displaystyle M \displaystyle= \displaystyle-\frac{d_{I}}{d_{O}}
\displaystyle= \displaystyle 0.65
\displaystyle d_{I} \displaystyle= \displaystyle-(0.65)(2.2\;\mathrm{m})
\displaystyle= \displaystyle-1.\underline{4}3\;\mathrm{m}

Now find the focal length.

\displaystyle\frac{1}{f} \displaystyle= \displaystyle\frac{1}{d_{O}}+\frac{1}{d_{I}}
\displaystyle= \displaystyle\frac{1}{3.0\;\mathrm{m}}+\frac{1}{-1.\underline{4}3\;\mathrm{m}}
\displaystyle f= \displaystyle-4.1\;\mathrm{m}

§ Chapter 23, Problem 27

A diver shines a flashlight upward from beneath the water at a 42.5^{0} angle to the vertical. At what angle does the light leave the water?

\displaystyle n_{1}\sin\theta _{1} \displaystyle= \displaystyle n_{2}\sin\theta _{2}
\displaystyle 1.33\sin 42.5^{0} \displaystyle= \displaystyle 1\,\sin\theta _{2}
\displaystyle\sin\theta _{2} \displaystyle= \displaystyle 0.899
\displaystyle\theta _{2} \displaystyle= \displaystyle\sin^{{-1}}(0.899)
\displaystyle= \displaystyle 64.0^{0}

§ Chapter 23, Problem 30

An aquarium filled with water has flat glass sides whose index of refraction is n_{\mathrm{glass}}=1.52 A beam of light from outside the aquarium strikes the glass at angle of 43.5^{0} to the perpendicular.

§ (a)

What is the angle of this light ray when it enters the glass?

\displaystyle n_{1}\sin\theta _{1} \displaystyle= \displaystyle n_{2}\sin\theta _{2}
\displaystyle 1\,\sin 43.5^{0} \displaystyle= \displaystyle 1.52\sin\theta _{2}
\displaystyle\sin\theta _{2} \displaystyle= \displaystyle 0.453
\displaystyle\theta _{2} \displaystyle= \displaystyle\sin^{{-1}}(0.453)
\displaystyle= \displaystyle 26.9^{0}

§ (b)

What is the angle of this light ray when it enters the water?

\displaystyle n_{3}\sin\theta _{3} \displaystyle= \displaystyle n_{2}\sin\theta _{2}
\displaystyle 1.33\sin\theta _{3} \displaystyle= \displaystyle 1.52\sin 26.9^{0}
\displaystyle\sin\theta _{3} \displaystyle= \displaystyle 0.518
\displaystyle\theta _{3} \displaystyle= \displaystyle\sin^{{-1}}(0.518)
\displaystyle= \displaystyle 31.2^{0}

Look at the diagram to see how the refracted angle from the first application of Snell’s Law becomes the incident angle for the second step.

§ (c)

What would be the angle of this light ray if it entered the water directly?

\displaystyle n_{3}\sin\theta _{3} \displaystyle= \displaystyle n_{1}\sin\theta _{1}
\displaystyle 1.33\sin\theta _{3} \displaystyle= \displaystyle 1\,\sin 43.5^{0}
\displaystyle\sin\theta _{3} \displaystyle= \displaystyle 0.518
\displaystyle\theta _{3} \displaystyle= \displaystyle\sin^{{-1}}(0.518)
\displaystyle= \displaystyle 31.2^{0}

§ Chapter 23, Problem 49

A lens with focal length f=80\;\mathrm{mm} is used to focus an image on the film of a camera. The maximum distance allowed between the lens and the film plane is 120\;\mathrm{mm}.

§ (a)

How far should the film be from the lens if the objected to be photographed is 10.0\;\mathrm{m} away?

\displaystyle\frac{1}{d_{O}}+\frac{1}{d_{I}} \displaystyle= \displaystyle\frac{1}{f}
\displaystyle\frac{1}{10.0\;\mathrm{m}}+\frac{1}{d_{I}} \displaystyle= \displaystyle\frac{1}{0.080\;\mathrm{m}}
\displaystyle d_{I} \displaystyle= \displaystyle 81\;\mathrm{mm}

Note that 10.0\;\mathrm{m} is pretty far away compared to the focal length, so the rays are coming in nearly parallel, and we expect the image to appear near f. That is where the film should be placed.

§ (b)

How far should the film be from the lens if the objected to be photographed is 3.0\;\mathrm{m} away?

\displaystyle\frac{1}{d_{O}}+\frac{1}{d_{I}} \displaystyle= \displaystyle\frac{1}{f}
\displaystyle\frac{1}{3.0\;\mathrm{m}}+\frac{1}{d_{I}} \displaystyle= \displaystyle\frac{1}{0.080\;\mathrm{m}}
\displaystyle d_{I} \displaystyle= \displaystyle 82\;\mathrm{mm}

Notice that the object is still a distance away that is 37.5\, f.

§ (c)

How far should the film be from the lens if the objected to be photographed is 1.0\;\mathrm{m} away?

\displaystyle\frac{1}{d_{O}}+\frac{1}{d_{I}} \displaystyle= \displaystyle\frac{1}{f}
\displaystyle\frac{1}{1.0\;\mathrm{m}}+\frac{1}{d_{I}} \displaystyle= \displaystyle\frac{1}{0.080\;\mathrm{m}}
\displaystyle d_{I} \displaystyle= \displaystyle 87\;\mathrm{mm}

§ (d)

What is the closest image that can be photographed in focus?

\displaystyle\frac{1}{d_{O}}+\frac{1}{d_{I}} \displaystyle= \displaystyle\frac{1}{f}
\displaystyle\frac{1}{d_{O}}+\frac{1}{0.120\;\mathrm{m}} \displaystyle= \displaystyle\frac{1}{0.080\;\mathrm{m}}
\displaystyle d_{0} \displaystyle= \displaystyle 0.24\;\mathrm{m}

At this distance the film is fully 120\;\mathrm{mm} away from the lens, at its physical limit.

§ Chapter 23, Problem 52

How far from a converging lens should an object be placed to produce a real image which is the same size as the object?

From the magnification equation

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

note that the image will be inverted. Then

\displaystyle\frac{1}{d_{O}}+\frac{1}{d_{I}} \displaystyle= \displaystyle\frac{1}{f}
\displaystyle\frac{1}{d_{O}}+\frac{1}{d_{O}} \displaystyle= \displaystyle\frac{1}{f}
\displaystyle\frac{2}{d_{O}} \displaystyle= \displaystyle\frac{1}{f}
\displaystyle d_{O} \displaystyle= \displaystyle 2f

§ Chapter 23, Problem 53

§ (a)

How far from a lens with a 50.0\;\mathrm{mm} focal length must an object be placed if its image is to be magnified 2.00\times and be real?

\displaystyle M \displaystyle= \displaystyle-\frac{d_{I}}{d_{O}}
\displaystyle= \displaystyle-2.00
\displaystyle d_{I} \displaystyle= \displaystyle 2.00\, d_{O}

Then

\displaystyle\frac{1}{d_{O}}+\frac{1}{d_{I}} \displaystyle= \displaystyle\frac{1}{f}
\displaystyle\frac{1}{d_{O}}+\frac{1}{2.00\, d_{O}} \displaystyle= \displaystyle\frac{1}{f}
\displaystyle d_{O} \displaystyle= \displaystyle 1.50\, f
\displaystyle= \displaystyle 75.0\;\mathrm{mm}

§ (b)

How far from a lens with a 50.0\;\mathrm{mm} focal length must an object be placed if its image is to be magnified 2.00\times and be virtual?

\displaystyle M \displaystyle= \displaystyle-\frac{d_{I}}{d_{O}}
\displaystyle= \displaystyle 2.00
\displaystyle d_{I} \displaystyle= \displaystyle-2.00\, d_{O}

Then

\displaystyle\frac{1}{d_{O}}+\frac{1}{d_{I}} \displaystyle= \displaystyle\frac{1}{f}
\displaystyle\frac{1}{d_{O}}+\frac{1}{-2.00\, d_{O}} \displaystyle= \displaystyle\frac{1}{f}
\displaystyle d_{O} \displaystyle= \displaystyle 0.50\, f
\displaystyle= \displaystyle 25.0\;\mathrm{mm}

§ Chapter 23, Problem 55

§ (a)

An insect 2.00\;\mathrm{cm} high is 1.20\;\mathrm{m} from a lens with a 135\;\mathrm{mm} focal length. Where is the image, how high is it, and what type is it?

\displaystyle\frac{1}{d_{O}}+\frac{1}{d_{I}} \displaystyle= \displaystyle\frac{1}{f}
\displaystyle\frac{1}{1.20\;\mathrm{m}}+\frac{1}{d_{I}} \displaystyle= \displaystyle\frac{1}{0.135\;\mathrm{m}}
\displaystyle d_{I} \displaystyle= \displaystyle 0.152\;\mathrm{m}

This is a real image 152\;\mathrm{mm} from the lens, on the opposite side as the object.

\displaystyle M \displaystyle= \displaystyle-\frac{d_{I}}{d_{O}}
\displaystyle= \displaystyle-\frac{0.152\;\mathrm{m}}{1.20\;\mathrm{m}}
\displaystyle= \displaystyle-0.127
\displaystyle h_{I} \displaystyle= \displaystyle Mh_{O}
\displaystyle= \displaystyle(-0.127)(20.0\;\mathrm{mm})
\displaystyle= \displaystyle 2.54\;\mathrm{mm}

The image is inverted.

§ (b)

What if f=-135\;\mathrm{mm}?

\displaystyle\frac{1}{d_{O}}+\frac{1}{d_{I}} \displaystyle= \displaystyle\frac{1}{f}
\displaystyle\frac{1}{1.20\;\mathrm{m}}+\frac{1}{d_{I}} \displaystyle= \displaystyle\frac{1}{-0.135\;\mathrm{m}}
\displaystyle d_{I} \displaystyle= \displaystyle-0.121\;\mathrm{m}

This is a virtual image 121\;\mathrm{mm} in front of the lens, on the same side as the object.

\displaystyle M \displaystyle= \displaystyle-\frac{d_{I}}{d_{O}}
\displaystyle= \displaystyle-\frac{-0.121\;\mathrm{m}}{1.20\;\mathrm{m}}
\displaystyle= \displaystyle+0.101
\displaystyle h_{I} \displaystyle= \displaystyle Mh_{O}
\displaystyle= \displaystyle(+0.101)(20.0\;\mathrm{mm})
\displaystyle= \displaystyle 2.02\;\mathrm{mm}

The image is upright.

 

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