An object is placed 16.0 cm from a screen where it is tobe imaged. (a) At what p

Question

An object is placed 16.0 cm from a screen where it is tobe imaged.

(a) At what point or points (there may be more than onesolution) should a converging lens

with focal length 3.00 cm be placed so that it willproduce a sharp image on the screen?

(b) Draw a ray diagram for each case. Your ray diagram(s)should be semi-quantitative, i.e.,

make the drawings at least 5 or 6” in overall sizeand use a ruler to help draw the principle

rays.

(c)Describe the resulting image(s) and calculate the magnification foreach case.

Explanation / Answer

(a)point at which converging lens should be placed (1/f) = (1/u) + (1/v) or (1/v) = (1/f) - (1/u) = (1/3.00) - (1/16.0) = (16.0 -3.00/48) = (13/48) or v = (48/13) cm (c)Because the lens is converging,the focal length ispositive.We expect the possibilities of both real and virtualimages. Because the object is larger than the focal length,we expectthe image to be real.The positive sign for the image distance tellsus that the image is indeed real and on the back side of thelens.The magnification tells us that the image is reduced in heightby one half,and the negative sign for M tells us that the image isinverted. the magnification is M = (v/u) = ((48/13)/16.0) = (3/13) = 0.23

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