a) An object is placed in front of a convex mirror. Draw aconvex mirror (radius
ID: 1746124 • Letter: A
Question
a) An object is placed in front of a convex mirror. Draw aconvex mirror (radius of curvature = 15cm) to scale, and place an object 27 cm infront of it. Make the object height 4 cm. Using a ray diagram,locate the image and measure its height. (Do this diagram on aseparate piece of paper. You will NOT turn this in but doing itwill help you with the assignment, help you better understand thematerial, and prepare you for things you WILL have to do on theexam.)1 cm (image location, include sign)
2 cm (image height, include sign)
(e) What is the ratio of the image height when the objectdistance is 5 cm to that when the objectdistance is 27 cm?
h5 cm/h27 cm = 7 a) An object is placed in front of a convex mirror. Draw aconvex mirror (radius of curvature = 15cm) to scale, and place an object 27 cm infront of it. Make the object height 4 cm. Using a ray diagram,locate the image and measure its height. (Do this diagram on aseparate piece of paper. You will NOT turn this in but doing itwill help you with the assignment, help you better understand thematerial, and prepare you for things you WILL have to do on theexam.)
1 cm (image location, include sign)
2 cm (image height, include sign)
(e) What is the ratio of the image height when the objectdistance is 5 cm to that when the objectdistance is 27 cm?
h5 cm/h27 cm = 7
Explanation / Answer
An object is placed in front of a convex mirror. Draw a convexmirror (radius of curvature = 15 cm) toscale, and place an object 27 cm in frontof it. Make the object height 4 cm. Using a ray diagram, locate theimage and measure its height. (Do this diagram on a separate pieceof paper. You will NOT turn this in but doing it will help you withthe assignment, help you better understand the material, andprepare you for things you WILL have to do on the exam.)1 cm (image location, include sign)
2 cm (image height, include sign)
(e) What is the ratio of the image height when the objectdistance is 5 cm to that when the objectdistance is 27 cm?
h5 cm/h27 cm = 7
R=15cm so f=7.5cm
u=-27cm
v=?
using 1/v-1/u=1/f
1/v=1/f+1/u=1/7.5+1/27=>1/v=0.1333+0.0370
v=5.871cm
v is positive hence the image is virtual
magnification=-v/u=-5.871/27= -0.2174 negative sign indicates thatthe image is virtual.
size of the image =MAGNIFICATION*SIZE OF THEOBJECT=0.2174*4=0.8697cm
for e follow the same procedure to findout image distance andmagnification and find out the ratio
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