LIGHT/OPTICS
Question:
A top view of a mirror and an arrow is shown below
a) Locate and sketch the image of the arrow
b) Position an observe's eye where the whole image could be seen
c) Draw a ray diagram that shows how light from both end of the arrow reach the observer
Diagram show below
A top view of a mirror and an arrow is shown below
a) Locate and sketch the image of the arrow
b) Position an observe's eye where the whole image could be seen
c) Draw a ray diagram that shows how light from both end of the arrow reach the observer
Diagram show below
Question:
1) a. Make a ray diagram to locate the image formed by the 0.92 m long baseball bat.
b. The bat is 5.0 m from the lens and the image formed is 0.25 m tall. Use similar triangles within your diagram or the thin lens equation to calculate the focal length of the lens.
c. Draw an eye at a position that could see the image of the bat.
Solution to b: To find the focal length of the lens, we can use the thin lens equation which is [(1/f) = (1/si)/(1/so)]
Because Si/So = Hi/Ho , we can calculate Si using cross multiplication because we are given the other 3 values.
Hi = .25 m
Ho = .92 m
So = 5.0 m
Si can be calculated to be 1.36 m.
Then we can plug in Si into the thin lens equation to find 'f'.
(1/f)= (1/1.36) + (1/5)
f can be calculated using algebra to be 1.07m. This is the distance of the lens' focal length.
Diagram show below
1) a. Make a ray diagram to locate the image formed by the 0.92 m long baseball bat.
b. The bat is 5.0 m from the lens and the image formed is 0.25 m tall. Use similar triangles within your diagram or the thin lens equation to calculate the focal length of the lens.
c. Draw an eye at a position that could see the image of the bat.
Solution to b: To find the focal length of the lens, we can use the thin lens equation which is [(1/f) = (1/si)/(1/so)]
Because Si/So = Hi/Ho , we can calculate Si using cross multiplication because we are given the other 3 values.
Hi = .25 m
Ho = .92 m
So = 5.0 m
Si can be calculated to be 1.36 m.
Then we can plug in Si into the thin lens equation to find 'f'.
(1/f)= (1/1.36) + (1/5)
f can be calculated using algebra to be 1.07m. This is the distance of the lens' focal length.
Diagram show below
Question:
Quantitatively determine the direction of the refracted ray.
Quantitatively determine the direction of the refracted ray.
Using Snells law, we can calculate theta(2)
Snells law is n1sin(theta1) = n2sin(theta2)
Given n1 = 1.3, theta1 = 40degrees, n2 = 1.5. So plugging into the equation above, we get:
1.3*sin(40) = 1.5*sin(theta2)
sin(theta2)=(1.3*sin40)/(1.5)
=.557
Therefore, theta2 = arcsin (.557)
=33.85 degrees
Snells law is n1sin(theta1) = n2sin(theta2)
Given n1 = 1.3, theta1 = 40degrees, n2 = 1.5. So plugging into the equation above, we get:
1.3*sin(40) = 1.5*sin(theta2)
sin(theta2)=(1.3*sin40)/(1.5)
=.557
Therefore, theta2 = arcsin (.557)
=33.85 degrees