Solve the simple lens/mirror equation for each situation to locate the image (fi

Question

Solve the simple lens/mirror equation for each situation to locate the image (find q) and calculate m. Where practical, do the calculation in unused space adjacent to the ray diagram. Set up a table to compare your graphical and calculated values for q and m. 1. An object 1.00" high is placed 3.50" in front of a concave mirror of radius 4.00". Locate and describe the image. 2. An object 1.00" high is placed .80" in front of a concave mirror of radius 4.00". Locate and describe the image. 3. An object 1.50" high is placed 2.00" in front of a convex mirror of radius 4.00". Locate and describe the image

Explanation / Answer

Given

   1. y = 1.0' , s = 3.5' , R = 4 ' ==> f = R/2 = 4/2 = 2 '

from the relation

   1/s + 1/s' = 1/f = 2/R

   1/s' = 1/2 - 1/3.5

   s' = 4.67 '

   m = y'/y = -s'/s ==> y' = -s'*y/s = -4.67*1/3.5 = -1.334

the image is inverted and enlarged real image

2.

   y = 1.0' , s = 0.8' , R = 4 ' ==> f = R/2 = 4/2 = 2 '

from the relation

   1/s + 1/s' = 1/f = 2/R

   1/s' = 1/2 - 1/0.8

   s' = -1.33 '

   m = y'/y = -s'/s ==> y' = -s'*y/s = 1.33*1/0.8 = 1.6625

the image is virtual enlarged image

3.

y = 1.5' , s= 2.0' , R = 4.0'

from the relation

   1/s + 1/s' = 1/f = 2/R

   1/s' = 1/2 - 1/2

   s'is at infinity

image will be formed at infinity only

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