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1. The radius of Earth is 6.40 x 10³ km. The moon is about 3.84 x 10^5 km away from Earth and has a diameter of 3475 km. The Pacific Ocean surface, which can be considered a convex mirror, forms a virtual image of the moon. What is the diameter of that image?
3. Ok. So since the image is virtual, q (image distance) is negative. p (object distance) is 3.84 x 10^5 km and h(object height) is 3475 m. Could the radius of Earth be used as the radius of curvature? I'm confused as to how to set up the equations in order to get the diameter of the image(h')
Code:
2. 1/p + 1/q = 1/f
Code:
M = h'/h = -q/p
3. Ok. So since the image is virtual, q (image distance) is negative. p (object distance) is 3.84 x 10^5 km and h(object height) is 3475 m. Could the radius of Earth be used as the radius of curvature? I'm confused as to how to set up the equations in order to get the diameter of the image(h')