1)the power of a lens is 2.5D.what is the focal length and type of lens. 2)the focal length of a concave mirror is 30cm.Find the position of the object in front of mirror so that the image is 3times the size of the object 3)an object is placed in front of a convex mirror of radius of curvature 60cm.Find the position of image and its magnification
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1. Power P = 1/f = 2.5D
f = 1/P = 100/2.5 = 40cm
Here f = 40cm which is positive, it means the lens is a convex lens.
2. In case of a concave f is negative. So in the question f = -30cm
Magnification m = -v/u = 3.
It means v = -3u.
Mirror formula = 1/f = 1/v + 1/u
1/ (-30) = 1/(-3u) + 1/u
-1/30 = 1/u – 1/3u
-1/30 = 3/(3u)
u = -20cms
It means the object is placed 20cms in front of the mirror. Since focal length is -30 cm ,it means the object is placed between focus and optical centre.
f = 1/P = 100/2.5 = 40cm
Here f = 40cm which is positive, it means the lens is a convex lens.
2. In case of a concave f is negative. So in the question f = -30cm
Magnification m = -v/u = 3.
It means v = -3u.
Mirror formula = 1/f = 1/v + 1/u
1/ (-30) = 1/(-3u) + 1/u
-1/30 = 1/u – 1/3u
-1/30 = 3/(3u)
u = -20cms
It means the object is placed 20cms in front of the mirror. Since focal length is -30 cm ,it means the object is placed between focus and optical centre.
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