Sam holds a lit match in front of a lens (f = 20.0 cm). The distance between the

Question

Sam holds a lit match in front of a lens (f = 20.0 cm). The distance between the match and the lens is S_o = 30.0 cm. Answer these three questions using three significant figures. A Pick the selection that describes the characteristics of the image from the match. real and upside down real and erect virtual and upside down virtual and erect What is the value of the image distance? s_i = If the flame is actually 1.500 cm tall, how tall is the image of the flame? (Enter your answer as a positive number.) h_i =

Explanation / Answer

A. When the object is located in front of the 2F point, the image will be located beyond the 2F point on the other side of the lens. Regardless of exactly where the object is located between 2F and F, the image will be located in the specified region. In this case, the image will be inverted (i.e., a right side up object results in an upside-down image). The image dimensions are larger than the object dimensions. A six-foot tall person would have an image that is larger than six feet tall. The absolute value of the magnification is greater than 1. Finally, the image is a real image.

So, Option 1 - real and upside down.

B. 1 / f = 1 / do + 1 / di

So,
1/20 = 1/30 + 1/di

di = 60 cm

C.
The magnification equation relates the ratio of the image distance and object distance to the ratio of the image height (hi) and object height (ho). The magnification equation is stated as follows:

M = hi/ho = -di/do

So,

hi/1.5 = -60/30

or hi = -2 * 1.5
= -3 cm

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