Asked by Alicia
An object 6.70 cm high is placed 15.1 cm in front of a convex mirror with radius of curvature of 8.55 cm. Where is the image formed?
Answers
Answered by
Alicia
1/q + 1/p = 1/f f=r/2 p=15.1
f= 8.55/2= 4.275
1/q= 1/f+1/p= 1/4.275 + 1/15.1 = 19.375/64.5525
take inverse to get q
q= 3.33
f= 8.55/2= 4.275
1/q= 1/f+1/p= 1/4.275 + 1/15.1 = 19.375/64.5525
take inverse to get q
q= 3.33
Answered by
Elena
do=15.1 cm, R=8.55 cm, H= 6.7 cm
di= ? h=?
1/do – 1/di = -2/R
1/di = 1/do+2/R = 1/15.1 +2/8.55.
di= 3.33 cm
The virtual image is smaller and closer to the mirror than the object
h/di =H/do =>
h=H•di/do = 6.7•3.33/15.1 =1.48 cm
di= ? h=?
1/do – 1/di = -2/R
1/di = 1/do+2/R = 1/15.1 +2/8.55.
di= 3.33 cm
The virtual image is smaller and closer to the mirror than the object
h/di =H/do =>
h=H•di/do = 6.7•3.33/15.1 =1.48 cm
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