Part A Calculate the location of the image formed by an 8.00-mm-tall object whos
ID: 1985362 • Letter: P
Question
Part ACalculate the location of the image formed by an 8.00-mm-tall object whose distance from the mirror is 15.0cm .
=-3.75 cm
Correct
Part B
Calculate the size of the image.
=2.00 mm
Correct
Part C
Calculate the location of the image formed by an 8.00-mm-tall object whose distance from the mirror is 10.0 cm .
=-3.33 cm
Correct
Part D
Calculate the size of the image.
=2.67 mm
Correct
Part E
Calculate the location of the image formed by an 8.00-mm-tall object whose distance from the mirror is 2.50 cm.
=-1.67 cm
Correct
Part F
Calculate the size of the image.
= mm
Part G
Calculate the location of the image formed by an 8.00-mm-tall object whose distance from the mirror is 10.0 cm .
=-4.98 cm
Correct
Part H
Calculate the size of the image.
= mm
Explanation / Answer
As the mirror is convex the focal length is negative
We have f=R/2 = 10.cm/2 = -5.0cm
a)
The object distance, image distance and focal length are related by
1/f = 1/s+1/s'
1/s' = 1/f-1/s
1/s' = s-f/fs
or s' = fs/s-f
= (-5.0cm)(15.0cm)/15cm+5cm
= -3.75 cm
we have magnification as M= y'/y = -s'/s
or y'=(-s'/s)*y
= (3.75cm/15cm)*8mm
= 2.0mm
b)
we have
s' = fs/s-f
= (-5.0cm)(10.0cm)/10cm+5cm
= -3.33 cm
we have magnification as M= y'/y = -s'/s
or y'=(-s'/s)*y
= (3.33m/10cm)*8mm
= 2.67mm
c)
we have
s' = fs/s-f
= (-5.0cm)(2.5cm)/2.5cm+5cm
= -1.67 cm
we have magnification as M= y'/y = -s'/s
or y'=(-s'/s)*y
= (1.67m/2.5cm)*8mm
= 5.344mm
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.