1) A person walks into a room that has, on opposite walls, two plane mirrors pro

Question

1) A person walks into a room that has, on opposite walls, two plane mirrors producing multiple images. Find the distances from the person to the first three images seen in the left-hand mirror, when the person is 5.00 ft from the mirror on the left wall and 8.00 ft from the mirror on the right wall.

a) Frist image in ft

b) Second image in ft

c) Third image in ft

2) At an intersection of hospital hallways, a convex mirror is mounted high on a wall to help people avoid collisions. The mirror has a radius of curvature of 0.530 m. Locate the image of a patient 10.0 m from the mirror.
a) q = in m

b) Is it in: front of the mirror / behind the mirror

c) Determine the magnification of the image.
M =  

d) Describe the image. (Select all that apply.)

upright

inverted

virtual

real

embiggened

minified

Explanation / Answer

1)
For left mirror, 1st image will be formed at 5 ft from left mirror

for right mirror, 1st image will be formed at 8 ft from right mirror which is at a distance of 13+8 = 21 ft from left mirror

So, 2nd image by left mirror is formed ar 21 ft
SImilarly 3rd image if formed at: 5+13+13 = 31 ft

Answers are:
5 ft
21 ft
31 ft

2)
R = - 0.53 m
f = R/2
= -0.53/2
= -0.265 m

object distance, p = - 10 m
image distance is q

use:
1/f = 1/p + 1/q
1/(-0.265) = 1/(-10) + 1/q
1/q = 1/10 - 1/(0.265)
q= -0.272 m
negative sign shows image is on the same side as object
a)
q = 0.272 m
b)
front of mirror
c)
M = -q/p
= - (-0.272)/(-10)
= -0.0272
Negative sign shows image is inverted
Answer: 0.0272

d)
inverted and smaller and real

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