Assignment 18 Query

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course Phy 122

3/25 10:30 am

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:

I am a little confused at how to do this.

I know that we have to use sin(angle) in the equation somewhere, but I am struggling with the concepts surrounding refraction in water. If maybe you could try to explain it on super simple terms (I am not very good at understanding physics) I would greatly appreciate it! Thank you!

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Snell's Law says that

sin(theta_incident) / sin(theta_refracted) = n2 / n1,

where n1 is the index of refraction in the 'incident' material and n2 the index of refraction in the 'refracting' material, and theta_incident and theta_refracted are the incident and refracted angles, respectively..

Note that the sine of the incident angle is in the numerator on the left-hand side, while the index of refraction of the 'refracting' material is in the numerator on the right.

The incident and refracted angles are measured with respect to the normal direction, which is the direction perpendicular to the boundary between the materials.

In the present case the 'incident' material is the water, since that's where the beam is initially, and the 'refracting' material is air. The water surface is horizontal, so that the vertical is the normal direction.

So you know that theta_refracted = 66 degrees. You also know or can find the index of refraction of air (which is close to 1) and water (which is about 1.33).

So you know three of the four quantities present in the equation

sin(theta_incident) / sin(theta_refracted) = n2 / n1,

You can therefore proceed to solve for the fourth quantity, which is theta_incident.

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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):

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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:

If I understand this right, we take 1/0.205 m = 4.88 m^-1 = 4.88 diopters

Since this is positive, it is converging.

Now we take 1/-6.25 m^-1 = -0.16 m

Since it is negative, it is diverging.

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