use ray tracing to lovate the image formed by concave mirror with focal length of 7.50 cm if the object is placed 10.0 cm in front of the mirror. Is the image upright or inverted? Use object height= 1cm the attached photo is the example.

College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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use ray tracing to lovate the image formed by concave mirror with focal length of 7.50 cm if the object is placed 10.0 cm in front of the mirror. Is the image upright or inverted? Use object height= 1cm the attached photo is the example.
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Examples
A concave mirror forms an image, on a wall 3.00 m in front of the
mirror, of a headlamp filament 10.0 cm in front of the mirror. the mirror to the object (s) and from the mirror to the i
(a) What are the radius of curvature and focal length of the mirror? We solve Eq. (34.4) for R. and then use Eq. (34.5) to find f
(b) What is the lateral magnification? What is the image height if tion (34.7) yields both m and y'.
the object height is 5.00 mm?
nification m, and image height y'. We are given the dist
EXECUTE: (a) Both the object and the image are on the concave
(reflective) side of the mirror, so both s and' are positive; we
have s= 10.0 cm and s'= 300 cm. We solve Eq. (34.4) for R:
SOLUTION
IDENTIFY and SET UP: Figure 34.15 shows our sketch. Our target
variables are the radius of curvature R. focal length f, lateral mag-
2
10.0 cm
300 cm
R
R= 2
- 194 cm
34.15 Our sketch for this problem.
10.0 cm
300 cm
The focal length of the miror is f= R/2= 9.7 cin.
(b) From Eq. (34,7) the lateral magnification is
Ocrven
Object
h=500 mm
Mitor
300 cm
-30,0
Imge
Optle ale
10.0 cm
Because m is tiegative, the image is inverted. The height of the
image is 30.0 times the height of the object, or (30.0)(5.00 mm)
150 mm
s'= 300 m
Transcribed Image Text:ED Page view A Read aloud Add text V Draw Highlight Erase Examples A concave mirror forms an image, on a wall 3.00 m in front of the mirror, of a headlamp filament 10.0 cm in front of the mirror. the mirror to the object (s) and from the mirror to the i (a) What are the radius of curvature and focal length of the mirror? We solve Eq. (34.4) for R. and then use Eq. (34.5) to find f (b) What is the lateral magnification? What is the image height if tion (34.7) yields both m and y'. the object height is 5.00 mm? nification m, and image height y'. We are given the dist EXECUTE: (a) Both the object and the image are on the concave (reflective) side of the mirror, so both s and' are positive; we have s= 10.0 cm and s'= 300 cm. We solve Eq. (34.4) for R: SOLUTION IDENTIFY and SET UP: Figure 34.15 shows our sketch. Our target variables are the radius of curvature R. focal length f, lateral mag- 2 10.0 cm 300 cm R R= 2 - 194 cm 34.15 Our sketch for this problem. 10.0 cm 300 cm The focal length of the miror is f= R/2= 9.7 cin. (b) From Eq. (34,7) the lateral magnification is Ocrven Object h=500 mm Mitor 300 cm -30,0 Imge Optle ale 10.0 cm Because m is tiegative, the image is inverted. The height of the image is 30.0 times the height of the object, or (30.0)(5.00 mm) 150 mm s'= 300 m
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