Query 18

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course Phy 122

018. `Query 16

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Question: `qPrinciples of Physics and General College Physics 23.28 A light beam exits the water surface at 66 degrees to vertical. At what angle was it incident from under the water?

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Your solution:

Since we know that the angle of refraction will be less than the angle of incident, we start with the indices of refraction table on page 656.

Waters index of refraction is 1.33 so, working from the parallel:

Sin(1) / Sin(1.33) = .75

.75*sin(66) = .75*.9 = .69

Sin^-1(66) = 43.6 degrees

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Your solution is good, as is your reasoning, but your notation isn't correct. You want to clean that up to avoid potential confusion.

sin(1) and sin(1.33), as well as sin^-1(66), are not quantities associated with this problem sin^-1(66) would not be associated with any problem, since sin^-1 can only be applied to numbers between -1 and 1.

Using theta_i for the angle of incidence your first equation would have been

sin(theta_i) / sin(theta_r) = 1 / 1.33

which gives you

sin(theta_i) / sin(66 deg) = .75

It follows that sin(theta_i) = .75 * sin(66 deg) = .75 * .914 = .685.

Thus

theta_i = sin^-1(.685) = 43.2 degrees.

You were given the 66 degree angle with only 2 significant figures, so the appropriate answer would be

theta_i = 43 degrees.

Do note that the given solution says 'very approximately' after the 37 degrees. Given solutions are only guidelines. That said, 37 degrees is a lot further off than I prefer.

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confidence rating #$&*:

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Given Solution:

`a**** The angle of incidence and the angle of refraction are both measured relative to the normal direction, i.e., to the direction which is perpendicular to the surface. For a horizontal water surface, these angles will therefore be measured relative to the vertical.

66 degrees is therefore the angle of refraction.

Using 1.3 as the index of refraction of water and 1 as the index of refraction of air, we would get

• sin(angle of incidence) / sin(angle of refraction) = 1 / 1.3 = .7, very approximately, so that

• sin(angle of incidence) = 1.3 * sin(angle of refraction) = .7 * sin(66 deg) = .7 * .9 = .6, very approximately, so that

• angle of incidence = arcSin(.6) = 37 degrees, again a very approximate result.

You should of course use a more accurate value for the index of refraction and calculate your results to at least two significant figures.

STUDENT QUESTION

it should be 1/1.333 right? nb is where its going which is air

sin(66)/sin (theta)=1/1.333=.75

INSTRUCTOR RESPONSE

The incident beam is in the water (call this medium a, consistent with your usage), the refracted beam in the air (medium b).

It's sin(theta_a) / sin(theta_b) = n_b / n_a, so

sin(theta_a) = 1 / 1.333 * sin(theta_b) = .7 sin(66 dec) = .6 and

theta_a = 37 degrees.

Again all calculations are very approximate.

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Self-critique (if necessary): I went back to work the problem using 1.3 and rounding off like it appears was done in the solution. When I used the book, I am pretty sure that I did the problem correctly but came up with a difference of over 7 degrees. Is the way that I worked the problem correct or should I concentrate on the rounding to the ones, or tenths place with working problems like this in the future?

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Given solutions are guidelines.

Your solution should be accurate to within a number of significant figures consistent with the given information.

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Self-critique Rating:3

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Question: `qPrinciples of Physics and General College Physics 23.46 What is the power of a 20.5 cm lens? What is the focal length of a -6.25 diopter lens? Are these lenses converging or diverging?

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Your solution:

20.5cm = .205m so 1/.205 = 4.88, Positive so a converging lens.

1 / -6.25 = -.16, Negative so a diverging lens.

confidence rating #$&*:

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Given Solution:

`aThe power of the 20.5 cm lens is 1 / (.205 meters) = 4.87 m^-1 = 4.87 diopters.

A positive focal length implies a converging lens, so this lens is converging.

A lens with power -6.25 diopters has focal length 1 / (-6.25 m^-1) = -.16 m = -16 cm.

The negative focal length implies a diverging lens.

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Self-critique (if necessary):NA

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Self-critique Rating:NA

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&#Your work looks good. See my notes. Let me know if you have any questions. &#