In both situations the photo is erect, virtual, diminished. Only distinction is the in instance convex winter the picture is developed behind the whereas in instance of diverging lens the picture is created on the same side that the lens together the thing is.
You are watching: If an object is placed in front of a convex mirror, then the object will be
For an item to be put in front of a convex winter there space six distinctive positions. These are the corresponding images developed are thorough below:
Let F it is in Focal Point, C its facility of Curvature wherein C=2F
Object is placed between F and the mirror
Object is placed at the focal Point
Object is placed in between F and C
Object at C
Object past center that curvature
Object at infinity
As shown in the picture, beam diagram for a specific location all three rays the light starting from the peak of the object have to be expanded behind the mirror to discover the position of the pointer of image.Image in all six instances is erect, virtual, diminished, behind the mirror and between mirror and also F. Just for the sixth location it is situated at F and highly diminished to the size of a point.We can likewise use the mirror equation
#1/"Object street O"+1/"Image distance I"=1/("Focus "f)#
Sign convention is that #"O"# and #"I"# are optimistic if in front of the mirror, #f# is an unfavorable for convex mirror and #"I"# is an unfavorable for digital image.
Now we take the case of object placed in former of a diverging lens, the is a concave lens.
The ray diagram is as shown in the photo below for among the six locations of the object.
*Actually only any type of two out of these 3 are enough to acquire the place of the image.To achieve the photo the rays must be extended earlier a shown. Image in all six instances is erect, virtual, diminished, top top the very same side the the lens together the object is and also between lens and F. Only for the sixth location it is located at F and highly decreased to the dimension of a point.
We can likewise use the Lens equation for diverging lens#1/"Object distance O"+1/"Image distance I"=1/("Focus "f)#And straight magnification #M# equation;
#M-="Size that Image"/"Size that Object"=(-"I")/"O"#
Given the focal size is very same for both convex mirror, diverging lens, thus we acquire same result.
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Except the in situation mirror the picture is created behind the whereas in case of diverging lens the image is formed on the exact same side the the lens as the thing is.