1. If you place a 4.0cm high luminous object 45cm in front of a concave mirror with a focal length of 15cm, determine
a. where must you place a screen so as to have a clear image of the object that others can see without looking in the mirror.
b. the orientation of the image (upright or inverted).
c. the height of the image.
The worksheet gives us the diagram below. I have included the height of the object (given in the directions), as well as the locations of c and f. It tells us that the focal length is 15cm which means that the focus is 15cm in front of the mirror (the picture is drawn to a scale of each increment equaling 5cm, so therefore it is 3 increments in front of the mirror in my picture). The focus is always located 1/2 of the distance between the mirror and the center, so we know that the center is twice as far from the mirror as the focus is. Therefore, it is 30cm (represented by 6 increments in the diagram).
In order to complete the ray diagram, we will use the three rays that do not require a protractor. This will show us where the image will be projected. The first ray goes through the center (c). The rule is that this ray reflects as itself, so along the same line. The second ray is drawn parallel to the surface and reflects from the mirror through the focus (f). The third ray is drawn through the focus and reflects parallel to the surface. The point that these lines intersect at is the top of where the image is reflected since we drew our lines starting from the top of the original object. The bottom of the object is the surface. We connect these two points in order to draw the reflected image.
a. In order to find where a screen must be placed to have a clear image of the object that others can see without looking in the mirror, we must find the distance in front of the mirror that the image will appear. We use the formula one divided by focus equals one divided by object distance plus one divided by image distance (1 / f = 1 / object distance + 1 / image distance). The problem tells us that the distance from the object to the mirror is 45cm and the focus is 15cm in front of the mirror, so we plug those in. (Note: Object distance is abbreviated as So and image distance is abbreviated as Si.)
1 / f = 1 / So + 1 / Si
1 / 15 = 1 / 45 + 1 / Si
3 / 45 = 1 / 45 + 1 / Si
2 / 45 = 1 / Si
2Si = 45
Si = 45 / 2 or 22.5cm
b. To figure out the orientation of the image, meaning whether the image is upside down or right side up, we can simply look at our ray diagram. Since the reflected image that we drew is upside down compared to the original object, we can say that the image is inverted (simply meaning upside down). Another way to know is by using the work that we will do in part c (see below).
c. To find the height of the image, we can use a proportional formula: negative image distance divided by object distance equals image height divided by object height (-Si / So = hi / ho). We found the image distance in part a (22.5cm) and were given the object distance (45cm) and object height (4.0cm), so we merely plug that in to our formula.
-Si / So = hi / ho
-22.5cm / 45cm = hi / 4.0cm
-90cm = 45cm * hi
hi = -2cm
We got -2cm as the height of the image, yet we drop the negative sign since height cannot technically be less than zero. In the end, our image height equals 2cm.
As I said in part b, we can use the hi that we solved for to tell us whether or not the image is upright. In order to do this, we simply look at the sign. Since hi is -2cm, we know that the image must be inverted (upside down) because a negative height means that it is below the surface level.
Particle Model of Light Worksheet 7: Refraction
2. The light source is now under water. Sketch the path of the rays as they pass from the water into the air.
The above picture is what is given to us. In order to find where the light will refract, we must use Snell's law. The formula for this is n1 * sinθ1 = n2 * sinθ2. (Note: n stands for the indices of refraction, which are constants). The n of air is 1.0 and the n of water is 1.33, so we can plug those in. In order to find θ, we can draw a normal line (line perpendicular to the surface) to the arrow furthest to the right. We then measure the angle between that line and the arrow, which is 60°. Now we can plug them in to the formula.
n1 * sinθ1 = n2 * sinθ2
1.0 * sinθ1 = 1.33 * sin60°
sinθ1 = 1.151813787
arcsin(1.151813787) = dne
Once we find what sinθ1 equals, we use the inverse sine (or arcsin) of the number to find the angle. However, we then find out that the angle does not exist. In this case, we are certain that this is an instance of Total Internal Reflection (TIR). This can happen when light goes from a medium with a bigger index to one with a smaller index. The light literally bends so far that it doesn't pass the boundary between the two media (in this case, air and water). Therefore, it reflects at the same angle that it originally was at, so 60°.
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