Reflection: If you place a 4.0 cm high luminous object 45 cm in front of a concave mirror with a focal length of 15 cm, determine: a) where you must place a screen so as to have a clear image of the object that others can see without looking at the mirror b) the height of the image and c) the orientation of the image (upright or inverted).
We can first draw out the rays of light on the diagram to help us place the object, and then we can also prove all the answers mathematically. For the ray diagram, we can use the 3 special rays to help locate the image formed. The first special ray goes parallel to the axis and then reflects back through the focal point. The second special ray passes through the focal point and reflects parallel to the axis. The final special ray passes through the center and is reflected back through the center. The completed ray diagram is pictured below along with the location of the image.
To answer the questions mathematically, we can use the curved mirror equation and the magnification equation. First , we can calculate the distance at which the image will appear.
1/So + 1/Si = 1/f
1/0.45 m + 1/Si = 1/0.15
Si = 0.225 m or 22.5 cm
In order to see the image on a screen, the screen must be placed 22.5 cm in front of the mirror.
1/0.45 m + 1/Si = 1/0.15
Si = 0.225 m or 22.5 cm
In order to see the image on a screen, the screen must be placed 22.5 cm in front of the mirror.
To calculate the height of the image, we can use the magnification ratio. We know that magnification is defined as the height of the image divided by the height of the object. Magnification is also defined as (- image distance) divided by object distance. We know 3 out of the 4 variables, the unknown being the height of the image. Hence, we can easily solve for it.
-Si/So = Hi/Ho
-22.5 cm/45 cm = Hi/ 4.0cm
Hi = -2.0 cm
The image is 2 cm tall.
Because the image height is negative, this means that the object is in the opposite direction of the original object, meaning that it is inverted. The ray diagram also confirms this fact.
-22.5 cm/45 cm = Hi/ 4.0cm
Hi = -2.0 cm
The image is 2 cm tall.
Because the image height is negative, this means that the object is in the opposite direction of the original object, meaning that it is inverted. The ray diagram also confirms this fact.
Refraction: In a lab experiment where light passes from air into a plastic block, the incident angle is measured to be 25 degrees and the refracted angle is 21 degrees. Find the index of refraction for the block.
The angles of refraction are related through Snell's Law, which states that the index of refraction of the incident surface multiplied by the sine of the incident angle equals the index of refraction of the refracted surface multiplied by the sine of the refracted angle. Using this relation, we can solve for the index of refraction for the plastic.
n1sinθ1 = n2sinθ2
1.00sin25 = nsin21
n = 1.179
Therefore, the index of refraction for the plastic is 1.179.
1.00sin25 = nsin21
n = 1.179
Therefore, the index of refraction for the plastic is 1.179.