(1)A spherical mirror is to be used to form, on a screen located 5.30 m from the
ID: 2054900 • Letter: #
Question
(1)A spherical mirror is to be used to form, on a screen located 5.30 m from the object, an image six times the size of the object.What should be the magnitude of its radius of curvature?
What should be the magnitude of the distance between the mirror and the object?
(2)A converging lens has a focal length of 19.4 cm. Give the image distance and magnification for the following object distances, using the correct sign conventions. Select all options that correctly describe the image. If the image is located at infinity, enter "infinity" for the image distance and for the magnification, and select "no image is formed".
(A) object distance = 39.5 cm
image distance=38.12cm
magnification =-.965
a. side of lens opposite the object
b. no image is formed
c. real
d. inverted
e. upright
f. virtual
g. same side of lens as the object
(B)object distance = 10.0 cm
image distance= -20.63
magnification= 2.06
a. side of lens opposite the object
b. no image is formed
c. real
d. inverted
e. upright
f. virtual
g. same side of lens as the object
FOR #2 I ACTUALLY JUST NEED HELP ON THE MULTIPLE CHOICE SECTION
Explanation / Answer
q/p=6
q=6p
q+p=10.6m=7p
p=10.6/7=1.51m
q=9.09.....magnitude of the distance between the mirror and the object
1/q+1/p=2/r
1/9.09+1/1.51=1/r =>1.295m...the magnitude of its radius of curvature
The lens equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f). The equation is stated as follows:
1/f=1/d0+1/di
The magnification equation relates the ratio of the image distance and object distance to the ratio of the image height (hi) and object height (ho). The magnification equation is stated as follows:
M=hi/h0=di/d0
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