\"color:rgb(51,51,51);font-family:arial, helvetica, clean, sans-serif;font-size:
ID: 2111198 • Letter: #
Question
"color:rgb(51,51,51);font-family:arial, helvetica, clean, sans-serif;font-size:13px;line-height:16px;background-color:rgb(244,249,255);">
An object is 6
"middle" src=
"http://session.masteringphysics.com/render?expr={ m+cm}"
class="done" style=
"color:rgb(51,51,51);font-family:arial, helvetica, clean, sans-serif;font-size:13px;line-height:16px;background-color:rgb(244,249,255);" /> in
front of a convex mirror with a focal length of
"http://session.masteringphysics.com/render?expr={ m+cm}"
class="done" style=
"color:rgb(51,51,51);font-family:arial, helvetica, clean, sans-serif;font-size:13px;line-height:16px;background-color:rgb(244,249,255);" />.
"color:rgb(51,51,51);font-family:arial, helvetica, clean, sans-serif;font-size:13px;line-height:16px;background-color:rgb(244,249,255);">
"line-height:16px;">A
Use
ray tracing to determine the location of the image. (this is supposed to be a number guys, not describing the location using words)
"color:rgb(51,51,51);font-family:arial, helvetica, clean, sans-serif;font-size:13px;line-height:16px;">
"color:rgb(51,51,51);font-family:arial, helvetica, clean, sans-serif;font-size:13px;line-height:16px;">
B
"color:rgb(51,51,51);font-family:arial, helvetica, clean, sans-serif;font-size:13px;line-height:16px;">Is
the image upright or inverted?
"color:rgb(51,51,51);font-family:arial, helvetica, clean, sans-serif;font-size:13px;line-height:16px;">
"color:rgb(51,51,51);font-family:arial, helvetica, clean, sans-serif;font-size:13px;line-height:16px;">
C
"color:rgb(51,51,51);font-family:arial, helvetica, clean, sans-serif;font-size:13px;line-height:16px;">Is
the image real or virtual?
Explanation / Answer
Ok, for this lens system, we need first to work with the first lens, then this lens will make an image, we will use this image as an object for the other lens, and we will find the final location of the image, so here we fo : Note, the lens are 40 cm apart each other, and the object is at 30 cm to the left of the first lens For the first lens: 1 / p + 1 / q = 1 / f p = object distance q = image distance 1 / 30 + 1 / q = 1 / 10 q = 15 cm So the first image is 15 cm to the right of the first lens, and this image is at : 25 cm to the left of the second lens, now, let's work with the image and the second lens : 1 / 25 + 1 / q = 1 / 8 q = 11.76 cm So, well, the final image will be 11.76 cm to the right of the second lens. Now, the second lens is on : x = +20 The coordinate of the image is = x = + 31.76 cm Second part : There is a new lens of x = 0 cm The object is on x = -50, so 50 cm to the left. The image must be on x = 31.76 = q 1 / 50 + 1 / 31.76 = 1 / f3 f3 = 19.423 cm That's it Hope that helps
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