The convex mirror is the outside of the spherical surface is the reflecting surface.
The characteristic of concave mirror to spread of rays or DIVERGEN
The cut point of focusing reflected rays mentioned FOCUS LENGTH
Special rays of convex Mirror
1. A ray parallel to the mirror axis of a concave mirror seems to come from the
focal point of a convex mirror
2. A ray that passes through the focal point of a concave mirror is reflected parallel
to the mirror axis
3. A ray that proceeds along a radius of the mirror is reflected back original path
nShaping its image if an object is placed in front of the convex mirror
The nature of image is virtual, erect and diminished
The mirror equations can be applied to either concave or convex mirrors
1/s + 1/s’ = 1/f or 1/s + 1/s’ = 2/R
Because f = ½ R
Where: s = the distance of object
s’ = the distance of image
f = the distance of the focal point
R = radius of curvature
Magnification (M) is define as ratio between distance of the image(s’) to distance of the object (s) or ratio between height of the image (h’) to height of the object (h)
In mathematically can be formulated as follows:
M = s’/s or M = h’/h
Example 1
Question
A convex mirror of focal point 6 cm is placed 4 cm from the vertex of the mirror. Locate and describe the image formed by the mirror.
Solution: recalling that the focal point is negative for convex mirror, we can substitute
f = - 6 cm and s = 4 cm into the mirror equation
1/s’ = 1/f – 1/s
= 1/-6 -1/4
= - 2/12 – 3/12
= - 5/12
S’ = 12/-5
= - 2,4 cm
Since the image distance is negative, we have confirmed the image is virtual. The nature is virtual, erect and diminished.
Example 2
Question
What is the focal point of concave mirror whose radius of curvature is 20 cm?
What are the nature and location of an image formed by the mirror if an object is placed 15 cm from the vertex of the mirror?
What are magnification and height of the image, if the height of object is 2 cm?
Solution: The focal point is positive for a concave mirror and given by
f = ½ R = ½ x 20 cm = 10 cm
The object distance (s) = 15 cm, and the image distance can be found by solving the mirror equation explicitly for (s’)
1/s + 1/s’ = 1/f
1/s’ = 1/f – 1/s
s’ = s.f /s-f
Substituting f = + 10 cm and s = + 15 cm, we obtain
s’ = (15 cm ) ( 10 cm )/ 15 cm – 10 cm
= + 30 cm
The positive sign for s’ verifies that the image is real. The nature is real, inverted and enlarged.
Magnification of the image can be found as follows,
M = s’/ s
= 30/10
= 3 times
The height of the image can be found as follows,
h’ = M x h
= 3 x 2 cm
= 6 cm
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