Photoelectric Effect
http://www.lon-capa.org/~mmp/kap28/PhotoEffect/photo.htm
More visual Photoelectric effect:
http://phet.colorado.edu/simulations/photoelectric/photoelectric.jnlp
Solutions:
2000B5
a)
i) 4.5 eV or 7.2 x 10-19 J
ii) 1.26 x 106 m/s
b) 183 nm
c) 5.56 x 1014 Hz
1980B3
a) The graph should be straight forward, but you SHOULD know that all four points cannot be on the same line!
b) 0.75 x 1014 Hz
c) 3.1 eV
d) Energy of electrons is 5 eV according to data, so the stopping potential will be 5 volts.
e) Charges move in circular paths when they are in a magnetic field perpendicular to their velocity. This is a possible reason for this behavior.
Page 855:
16. 4.14 x 1014 Hz
17. 401 nm
19. The shortest visible wavelength photons have energies of 3.1 eV (400 nm wavelength). Copper and Iron have work functions that are greater than this, so they will not eject photo electrons in visible light.
20.
a) 3.49 x 10-19 J or 2.179 eV.
b) 0.930 V
Page 885:
P32. 4.14 x 10-11 m (Don't worry about the longest wavelength.)
P33. 41,000 V
Tuesday, March 31, 2009
Monday, March 30, 2009
From Handout:
1.
a) 2.5 mm
b) wavelength = 385 nm, frequency = 6 x 1014 Hz
c) Since the wavelength decreases, and the spacing between maxima is proportional to wavelength, the space between the maxima will decrease.
1991B6.
a) 3.9 x 10-5 m
b) 9.6 x 10-3 m
Problems:
7. 8.5 x 10-5m
10. The important thing here is to notice that the ratio d/L is the same for both. You should obtain the equation that 2*lambda_1/y = m*lambda_2/2*y, with y the same for both since the question asks for the wavelength at which a minimum appears at the same location. Solving for a wavelength lambda_2 in the visible spectrum, lambda_2 = 613 nm
20. Solve for theta using sin(theta)= m*lambda/W, and then use the fact that L*tan(theta) = height above the center. The distance is 1.591 meters.
21. Notice that this says the distance between maxima, not minima. The path difference must be half a wavelength more than that required for the minimum. The total angle is 19.5 degrees across the central maximum, so the angle between the central maximum and the first minimum is 9.75 degrees. sin(9.75) = 1.5*lambda/W, so W = 5606 nm.
1.
a) 2.5 mm
b) wavelength = 385 nm, frequency = 6 x 1014 Hz
c) Since the wavelength decreases, and the spacing between maxima is proportional to wavelength, the space between the maxima will decrease.
1991B6.
a) 3.9 x 10-5 m
b) 9.6 x 10-3 m
Problems:
7. 8.5 x 10-5m
10. The important thing here is to notice that the ratio d/L is the same for both. You should obtain the equation that 2*lambda_1/y = m*lambda_2/2*y, with y the same for both since the question asks for the wavelength at which a minimum appears at the same location. Solving for a wavelength lambda_2 in the visible spectrum, lambda_2 = 613 nm
20. Solve for theta using sin(theta)= m*lambda/W, and then use the fact that L*tan(theta) = height above the center. The distance is 1.591 meters.
21. Notice that this says the distance between maxima, not minima. The path difference must be half a wavelength more than that required for the minimum. The total angle is 19.5 degrees across the central maximum, so the angle between the central maximum and the first minimum is 9.75 degrees. sin(9.75) = 1.5*lambda/W, so W = 5606 nm.
Friday, March 27, 2009
There's this thin film on my teeth...(Eww)
Hi everyone,
Here is the link to the physics lesson website - be sure to check out the lesson on thin film interference, as well as on diffraction.
Please also be aware that there will be construction Saturday on the 6 line downtown on the way to the REACH workshop at Baruch College. Please give yourself plenty of time to get there by 8:00 AM.
From handout:
4. thickness = 1.05 x 10-7 m
5. thickness = 96.1 nanometers
1984B5
a) 5 x 1014 Hz
b) 4.8 x 10-7 m
c) 1.2 x 10-7 m
d) 2.4 x 10-7 m
2000B5.
I apologize for not getting this solutions up quickly, but you must be aware that part (a) of this question is worth 8 of the 15 points. You need to make sure you draw BOTH angles of reflection, BOTH angles of refraction (after using Snell's Law), and include the angle of incidence (35 degrees) of the light ray upon hitting the bottom surface of the glass.
Parts b and c are, by comparison, relatively tame. I will post this ASAP.
38. wavelength = 643 nm, which is colored red.
40. 169 nm
43. 27*lambda/n = 2t, so t = 9045 nanometers, or 9.045 x 10-6 m
45. 2t = n*(640 nm/1.36), and 2t = (n + 1/2)*512 nm/1.36. Solving the system, t = 471 nanometers.
Here is the link to the physics lesson website - be sure to check out the lesson on thin film interference, as well as on diffraction.
Please also be aware that there will be construction Saturday on the 6 line downtown on the way to the REACH workshop at Baruch College. Please give yourself plenty of time to get there by 8:00 AM.
From handout:
4. thickness = 1.05 x 10-7 m
5. thickness = 96.1 nanometers
1984B5
a) 5 x 1014 Hz
b) 4.8 x 10-7 m
c) 1.2 x 10-7 m
d) 2.4 x 10-7 m
2000B5.
I apologize for not getting this solutions up quickly, but you must be aware that part (a) of this question is worth 8 of the 15 points. You need to make sure you draw BOTH angles of reflection, BOTH angles of refraction (after using Snell's Law), and include the angle of incidence (35 degrees) of the light ray upon hitting the bottom surface of the glass.
Parts b and c are, by comparison, relatively tame. I will post this ASAP.
38. wavelength = 643 nm, which is colored red.
40. 169 nm
43. 27*lambda/n = 2t, so t = 9045 nanometers, or 9.045 x 10-6 m
45. 2t = n*(640 nm/1.36), and 2t = (n + 1/2)*512 nm/1.36. Solving the system, t = 471 nanometers.
Wednesday, March 25, 2009
It's Critical that you get the right angle!
Solutions to the AP Problems:
79B6.
a) Remember that you MUST use the angles relative to the normal vector - n1 sin(t1) = n2 sin (t2)
n2 = 1.327
n = c/v so v = 2.26 x 108 m/s
b) If you try to use Snell's law at point Q, you will find you end up trying to take the inverse sine of a number greater than 1 - not possible. This means the ray will NOT refract, but instead reflect. This is total internal reflection. You can justify the fact that it reflects either by calculating the critical angle and showing that the 53 degrees is greater than the critical angle (49 degrees), or by showing that when refraction does not occur, reflection does.
c) If the lens was made of plastic in air, it would be a converging lens. Since it is the reverse, it will be a diverging lens, and the rays will spread out.
88B5.
a) The ray will be refracted slightly downwards from the horizontal dotted line.
b)
The ray will pass directly through the left interface because the incident angle is 0. At the right interface, the angle from the normal is 37 degrees. By Snell's law, the refracted ray will have an angle of 64.5 degrees.
Be careful though, as the question asks for the angle from the horizontal. To get this, you must subtract the 37 degrees from the 64.5 degrees, giving a final answer of 27.5 degrees.
c) Here you must assume that the incident angle of 37 degrees is the critical angle.
By Snell's law, theta_1 = 37, n2 = 1.0 (for air), and theta_2 = 90 since we are assuming total internal reflection occurs. Solving, n1 = 1.667
d) Here the light ray will again be refracted below the horizontal line, but the angle should not be as great as it was for part (a).
e) Using Snell's law again with n1 = 1.667, theta_1 = 37 degrees, and n2 = 1.33, we can solve for theta_2 = 49.0 degrees. You must again subtract 37 degrees from this value to obtain the angle from the horizontal of 12 degrees below.
93B4.
a) v = c/n = 1.875 x 108 m/s
b) There are two ways to do this - the fast way is to remember the equation from Monday that the wavelength in the material lambda' = lambda/n where lambda is the wavelength in a vacuum. 700 nm/1.5 = 467 nm.
We can also use c = f*lambda to find the frequency of the red light in a vacuum:
f = 3E8 m/s/(700 E-9 m) = 4.28 x 1014 Hz
The frequency stays the same when a wave changes from one material to another, so this is also the frequency in the glass. We also know the speed of red light in the glass from part a.
lambda = v/f = 1.875 x 108 m/s/(4.28 x 1014 Hz) = 438 nm (either answer would be acceptable)
c) Frequency can be found the same as before - make SURE before you use the v = f*lambda equation, you are using the speed and wavelength for the same medium!
f = 3E8 m/s/(700 E-9 m) = 4.28 x 1014 Hz
d) The key here is to notice that the incident rays have the SAME initial angle hitting the right side of the prism (30 degrees), as well as the same n2 = 1 (for air.) The difference is in n1.
From Snell's Law:
n1 sin(theta_1) = n2*sin(theta_2)
sin(theta_2) = n1*sin(theta_1)
Since n1 is greater for the blue light, theta_2 will be greater for the blue light. This means that the blue light will bend further away relative to the normal than the red.
e) In this case, it is a similar situation when the rays reach the right surface (the rays again do not refract at the left surface because the incident angle is zero.) The issue here, however, is that n1 is the same for both (n1 = 1), and theta_1 is again 30 degrees.
Snell's law again:
n1 sin(theta_1) = n2*sin(theta_2)
sin(theta_2) = 1*sin(theta_1)/n1
In this case, since n1 is greater for blue light, it will refract less
than with the red light. Thus, the blue light will bend above the red light, with both rays bending towards the normal.
#3.
b) Using the thin lens equation, 1/di = 1/f - 1/do, so di = 30 cm. This should roughly match your answer for part (a).
c) hi/ho = -di/do so hi = 2*5cm = 10 cm
d) This can be a tricky one. If you use what we came up with today, you will find that the image from the first lens is on the right side of the second lens. You can use the rules as we always do, except that the primary focus of the second lens is now on the same side as the object, so the lens will act like a diverging lens.
The slick way to get this is to notice that Ray 2 leaves the first lens parallel to the axis when it enters the second lens. This ray enters lens 2 parallel, so it is brought to the primary focus of lens 2. Ray 1, after passing through lens 1, passes through the center of lens 2, which means it will pass through undeflected. This will result in these two rays converging at a single point to the right of lens 2. This is the location of the image, roughly 7 cm to the right of lens 2.
e) The image is inverted and smaller than the original object.
79B6.
a) Remember that you MUST use the angles relative to the normal vector - n1 sin(t1) = n2 sin (t2)
n2 = 1.327
n = c/v so v = 2.26 x 108 m/s
b) If you try to use Snell's law at point Q, you will find you end up trying to take the inverse sine of a number greater than 1 - not possible. This means the ray will NOT refract, but instead reflect. This is total internal reflection. You can justify the fact that it reflects either by calculating the critical angle and showing that the 53 degrees is greater than the critical angle (49 degrees), or by showing that when refraction does not occur, reflection does.
c) If the lens was made of plastic in air, it would be a converging lens. Since it is the reverse, it will be a diverging lens, and the rays will spread out.
88B5.
a) The ray will be refracted slightly downwards from the horizontal dotted line.
b)
The ray will pass directly through the left interface because the incident angle is 0. At the right interface, the angle from the normal is 37 degrees. By Snell's law, the refracted ray will have an angle of 64.5 degrees.
Be careful though, as the question asks for the angle from the horizontal. To get this, you must subtract the 37 degrees from the 64.5 degrees, giving a final answer of 27.5 degrees.
c) Here you must assume that the incident angle of 37 degrees is the critical angle.
By Snell's law, theta_1 = 37, n2 = 1.0 (for air), and theta_2 = 90 since we are assuming total internal reflection occurs. Solving, n1 = 1.667
d) Here the light ray will again be refracted below the horizontal line, but the angle should not be as great as it was for part (a).
e) Using Snell's law again with n1 = 1.667, theta_1 = 37 degrees, and n2 = 1.33, we can solve for theta_2 = 49.0 degrees. You must again subtract 37 degrees from this value to obtain the angle from the horizontal of 12 degrees below.
93B4.
a) v = c/n = 1.875 x 108 m/s
b) There are two ways to do this - the fast way is to remember the equation from Monday that the wavelength in the material lambda' = lambda/n where lambda is the wavelength in a vacuum. 700 nm/1.5 = 467 nm.
We can also use c = f*lambda to find the frequency of the red light in a vacuum:
f = 3E8 m/s/(700 E-9 m) = 4.28 x 1014 Hz
The frequency stays the same when a wave changes from one material to another, so this is also the frequency in the glass. We also know the speed of red light in the glass from part a.
lambda = v/f = 1.875 x 108 m/s/(4.28 x 1014 Hz) = 438 nm (either answer would be acceptable)
c) Frequency can be found the same as before - make SURE before you use the v = f*lambda equation, you are using the speed and wavelength for the same medium!
f = 3E8 m/s/(700 E-9 m) = 4.28 x 1014 Hz
d) The key here is to notice that the incident rays have the SAME initial angle hitting the right side of the prism (30 degrees), as well as the same n2 = 1 (for air.) The difference is in n1.
From Snell's Law:
n1 sin(theta_1) = n2*sin(theta_2)
sin(theta_2) = n1*sin(theta_1)
Since n1 is greater for the blue light, theta_2 will be greater for the blue light. This means that the blue light will bend further away relative to the normal than the red.
e) In this case, it is a similar situation when the rays reach the right surface (the rays again do not refract at the left surface because the incident angle is zero.) The issue here, however, is that n1 is the same for both (n1 = 1), and theta_1 is again 30 degrees.
Snell's law again:
n1 sin(theta_1) = n2*sin(theta_2)
sin(theta_2) = 1*sin(theta_1)/n1
In this case, since n1 is greater for blue light, it will refract less
than with the red light. Thus, the blue light will bend above the red light, with both rays bending towards the normal.
#3.
b) Using the thin lens equation, 1/di = 1/f - 1/do, so di = 30 cm. This should roughly match your answer for part (a).
c) hi/ho = -di/do so hi = 2*5cm = 10 cm
d) This can be a tricky one. If you use what we came up with today, you will find that the image from the first lens is on the right side of the second lens. You can use the rules as we always do, except that the primary focus of the second lens is now on the same side as the object, so the lens will act like a diverging lens.
The slick way to get this is to notice that Ray 2 leaves the first lens parallel to the axis when it enters the second lens. This ray enters lens 2 parallel, so it is brought to the primary focus of lens 2. Ray 1, after passing through lens 1, passes through the center of lens 2, which means it will pass through undeflected. This will result in these two rays converging at a single point to the right of lens 2. This is the location of the image, roughly 7 cm to the right of lens 2.
e) The image is inverted and smaller than the original object.
Tuesday, March 24, 2009
Can you Lens me a dollar? Times are tough....
Multiple Choice problems from the handout:
1. B
2. E
3. C
4. D
Questions:
20. If the object distance is very large, then by the thin lens equation, 1/d0 approaches zero. This means the image distance is equal to the focal length. This is where the film must be placed to form an image.
Problems:
48b. 390 mm
50. Since the image is on the other side of the lens, it must be a real image. This means that the image distance is positive. Using the thin lens equation, the focal length must be 39.8 cm.
52. 72 cm on the same side of the lens as the stamp (virtual image, negative image distance). Magnification = -4
60.
a) 15.2 cm, real image (opposite side of the lens as object), height is -2.78 mm
b) -12.1 cm, virtual image (negative image distance), height is +2.22 mm
83. Since a diverging lens has a negative focal length, by the thin lens equation, 1/di can never be positive. If di is negative, the image is on the same side as the object, so it will be virtual. The only way for di to be positive is if d0 for the object is negative. This can occur, and we will discuss HOW it can occur in class tomorrow!
AP Problems:
1989B6.
a) real
b) di = 30 cm
c) M = -0.5
d) See the image to the left.
e) focal length decreases - by Snell's law, the rays will be bent more as they enter and exit the lens, so the place where the rays intersect the principal axis will be closer to the center of the lens.
1997B5.
a) converging - the image is located behind the lens, which means image distance is positive. This can only occur with a converging lens.
b) 1/f = 1/di + 1/do so f = 22.5 mm
c) The rays will probably intersect near the 90 mm mark, but the 22.5 mm focal length makes it difficult to get it exactly. Don't worry if it isn't exactly on the 90 mm mark.
d) The image is inverted, real, and larger than the object. You can also confirm the size change by noticing that M = -di/do = -90/30 = -3.
e) The resulting image should be inverted and to the left of the mirror.
1. B
2. E
3. C
4. D
Questions:
20. If the object distance is very large, then by the thin lens equation, 1/d0 approaches zero. This means the image distance is equal to the focal length. This is where the film must be placed to form an image.
Problems:
48b. 390 mm
50. Since the image is on the other side of the lens, it must be a real image. This means that the image distance is positive. Using the thin lens equation, the focal length must be 39.8 cm.
52. 72 cm on the same side of the lens as the stamp (virtual image, negative image distance). Magnification = -4
60.
a) 15.2 cm, real image (opposite side of the lens as object), height is -2.78 mm
b) -12.1 cm, virtual image (negative image distance), height is +2.22 mm
83. Since a diverging lens has a negative focal length, by the thin lens equation, 1/di can never be positive. If di is negative, the image is on the same side as the object, so it will be virtual. The only way for di to be positive is if d0 for the object is negative. This can occur, and we will discuss HOW it can occur in class tomorrow!
AP Problems:
1989B6.
a) real
b) di = 30 cm
c) M = -0.5
d) See the image to the left.
e) focal length decreases - by Snell's law, the rays will be bent more as they enter and exit the lens, so the place where the rays intersect the principal axis will be closer to the center of the lens.
1997B5.
a) converging - the image is located behind the lens, which means image distance is positive. This can only occur with a converging lens.
b) 1/f = 1/di + 1/do so f = 22.5 mm
c) The rays will probably intersect near the 90 mm mark, but the 22.5 mm focal length makes it difficult to get it exactly. Don't worry if it isn't exactly on the 90 mm mark.
d) The image is inverted, real, and larger than the object. You can also confirm the size change by noticing that M = -di/do = -90/30 = -3.
e) The resulting image should be inverted and to the left of the mirror.
Monday, March 23, 2009
Come on Ray, just give me the equation!
A reminder: Please try the simulations below. They will go a long way to giving you an intuitive understanding of how the light rays move in both the cases of refraction and reflection. If you want a preview of tomorrow's lesson, try the lenses in the bottom simulation.
Index of Refraction simulator: http://www.ps.missouri.edu/rickspage/refract/refraction.html
Site for Simulating Mirrors (lenses too, coming soon to a 6/7 period near you!)
http://www.surendranath.org/Applets/Optics/ReflRefrCurv/CurvSurfApplet.html
Question 15 - The fish is, in reality, lower in the aquarium than is seen by the eye in the picture. Remember that the light ray from the fish is bent away from the normal when it leaves the water. This means that the actual ray from the fish to the water's surface is tilted downwards to reduce the angle from the normal to the surface.
Problems:
12. image is virtual, upright, and reduced in size. It is located at 2.09 cm behind the surface of the ball. (di = -2.09 cm).
13. magnification is -di/d0, so you can find di using the magnification of 3 (positive because it is upright.) Solving for the focal length, and multiplying by 2, the radius R = 3.9 meters.
14. concave mirror, R = 5.66 cm
27. n = c/v = 1.310
34. The light emerges from the first interface at an angle of 27.7 degrees. By geometry, we can figure out the angle at which this hits the right side of the prism. A quadrilateral is formed by the top vertex of the prism, the point where the refracted ray hits the left side, the right side intersection point, and the intersection of the normal vectors. (See the picture below)
We can see that this ray hits the other surface of the prism at 33 degrees. By Snell's Law, the ray emerges at 54.3 degrees.
35. Remember to find the angle from the normal! Theta 1 = 64.3 degrees, n1 = 1, n2 = 1.33, so theta 2 = 42.6 degrees. 2.1 meters * tan (42.6 degrees) = 1.93 meters. The total distance from the wall is therefore 2.7 m + 1.93 meters = 4.63 meters.
Index of Refraction simulator: http://www.ps.missouri.edu/rickspage/refract/refraction.html
Site for Simulating Mirrors (lenses too, coming soon to a 6/7 period near you!)
http://www.surendranath.org/Applets/Optics/ReflRefrCurv/CurvSurfApplet.html
Question 15 - The fish is, in reality, lower in the aquarium than is seen by the eye in the picture. Remember that the light ray from the fish is bent away from the normal when it leaves the water. This means that the actual ray from the fish to the water's surface is tilted downwards to reduce the angle from the normal to the surface.
Problems:
12. image is virtual, upright, and reduced in size. It is located at 2.09 cm behind the surface of the ball. (di = -2.09 cm).
13. magnification is -di/d0, so you can find di using the magnification of 3 (positive because it is upright.) Solving for the focal length, and multiplying by 2, the radius R = 3.9 meters.
14. concave mirror, R = 5.66 cm
27. n = c/v = 1.310
34. The light emerges from the first interface at an angle of 27.7 degrees. By geometry, we can figure out the angle at which this hits the right side of the prism. A quadrilateral is formed by the top vertex of the prism, the point where the refracted ray hits the left side, the right side intersection point, and the intersection of the normal vectors. (See the picture below)
We can see that this ray hits the other surface of the prism at 33 degrees. By Snell's Law, the ray emerges at 54.3 degrees.
35. Remember to find the angle from the normal! Theta 1 = 64.3 degrees, n1 = 1, n2 = 1.33, so theta 2 = 42.6 degrees. 2.1 meters * tan (42.6 degrees) = 1.93 meters. The total distance from the wall is therefore 2.7 m + 1.93 meters = 4.63 meters.
Sunday, March 22, 2009
Can YOU tell me what's on the other side of the mirror?
Hi everyone,
In case you had trouble watching the lesson in class Thursday and Friday, here is the site:
http://www.hippocampus.org/AP%20Physics%20B%20II
Click on 'Flat and Concave mirrors" under 'Geometric Optics' in the menu. You can also watch the Refraction video there as well.
Don't forget - we will have a ray tracing quiz and a problem from the handout. Feel free to check with me or each other to make sure you agree on the answers. Thanks to all of you who have already done this!
In case you had trouble watching the lesson in class Thursday and Friday, here is the site:
http://www.hippocampus.org/AP%20Physics%20B%20II
Click on 'Flat and Concave mirrors" under 'Geometric Optics' in the menu. You can also watch the Refraction video there as well.
Don't forget - we will have a ray tracing quiz and a problem from the handout. Feel free to check with me or each other to make sure you agree on the answers. Thanks to all of you who have already done this!
Subscribe to:
Posts (Atom)