An object of size 2.0 cm is placed perpendicular to the principal axis...
The size of the image will be 2.0 cm.
V=U(given)
m=-v/u
m= - 1
m=h`/h
-1=h`/2
h`=-2
h=2cm
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An object of size 2.0 cm is placed perpendicular to the principal axis...
Explanation:
Given, size of object (h1) = 2.0 cm, distance of object (u) = radius of curvature (R)
Using mirror formula, 1/f = 1/u + 1/v where f is the focal length and v is the distance of image from the mirror
As per the question, u = R, so 1/f = 1/R + 1/v
As per mirror formula, magnification (m) = -v/u = -v/R
Now, we can find the focal length using mirror formula and substituting u = R, we get
1/f = 1/R + 1/v
1/f = 2/R
f = R/2
Substituting this value of f in mirror formula, we get
1/u + 1/v = 1/f
1/R + 1/v = 2/R
1/v = 1/R
v = R
Now, we can find the size of the image using the magnification formula
m = -v/u = -R/R = -1
Size of image (h2) = |m| * h1 = 1 * 2.0 cm = 2.0 cm
Therefore, the correct option is D) 2.0 cm.
An object of size 2.0 cm is placed perpendicular to the principal axis...
It is clearly mentioned in the question that the distance of the object is equal to the radius of curvature of the concave mirror.
In this case the image formed is of same height as that of object.
So, height of the object =height of an image
=2cm
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