Problem 24.05 A concave spherical mirror has a radius of curvature of 14.0 cm .
ID: 1401929 • Letter: P
Question
Problem 24.05
A concave spherical mirror has a radius of curvature of 14.0 cm .
Part A
Calculate the location of the image formed by an 15.0-mm-tall object whose distance from the mirror is 28.0 cm .
9.33
Correct
Part B
Calculate the size of the image.
5.00
Correct
Part C
Calculate the location of the image formed by an 15.0-mm-tall object whose distance from the mirror is 14.0 cm .
14.0
Part D
Calculate the size of the image.
15.0
Correct
Part E
Calculate the location of the image formed by an 15.0-mm-tall object whose distance from the mirror is 3.50 cm .
-7.00
Correct
Part F
Calculate the size of the image.
30.0
Correct
Part G
Calculate the location of the image formed by an 15.0-mm-tall object whose distance from the mirror is 10.0 m .
Part H
Calculate the size of the image.
s1 =9.33
cmExplanation / Answer
A) focal length, f1 = R/2 = 14/2 = 7 cm
object distance, s = 28 cm
Apply, 1/s1 + 1/s1' = 1/f
1/s1' = 1/f - 1/s1
1/s1' = 1/7 - 1/28
s1' = 9.33 cm <<<<<<---------Answer
B) magnificationm, m = -s1'/s1
= -9.33/28
= -0.333
magnification, m = image height/object height
= y'/y
y' = m*y
= -0.333*15 mm
= -5 mm (here negative sign indicates image is inverted)
C)
1/s2' = 1/f - 1/s2
1/s2' = 1/7 - 1/14
s2' = 14 cm
D) magnificationm, m = -s2'/s2
= -14/14
= -1
magnification, m = image height/object height
= y'/y
y' = m*y
= -1*15 mm
= -15 mm (here negative sign indicates image is inverted)
E)
1/s3' = 1/f - 1/s3
1/s3' = 1/7 - 1/3.5
s3' = -7 cm <<<<<<---------Answer
F) magnificationm, m = -s3'/s3
= -(-7)/3.5
= 2
magnification, m = image height/object height
= y'/y
y' = m*y
= 2*15 mm
= +30 mm (here POSItive sign indicates image is UPRIGHT)
g)
1/s4' = 1/f - 1/s4
1/s4' = 1/7 - 1/1000
s3' = 7.05 cm <<<<<<---------Answer
h) magnificationm, m = -s4'/s4
= -7.05/1000
= -7.05*10^-3
magnification, m = image height/object height
= y'/y
y' = m*y
= -7.05*10^-3*15 mm
= -0.105 mm (here negative sign indicates image is inverted)
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