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प्रश्न
Find the position and magnification of the image of an object placed at distance of 8.0 cm in front of a convex lens of focal length 10.0 cm. Is the image erect or inverted?
उत्तर
Object distance=u=-8cm
Focal length f = 10 cm
Image distance v=?
`1/"v" - 1/"u" = 1/"f"`
`1/"v" - 1/-8 = 1/10`
`1/"v" = 1/10 - 1/8`
`1/"v" = (4 - 5)/40`
`1/"v" = 1/-40`
v = - 40 cm
As the object is placed between the focus and optical center of the lens, the image formed is virtual and erect.
Magnification (m) = `v/u = (-0)/-8`
m = +5.0
The +ve sign shows that the image formed is erect.
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संबंधित प्रश्न
A student is using a convex lens of focal length 10 cm to study the image formation by a convex lens for the various positions of the object. In one of his observations, he may observe that when the object is placed at a distance of 20 cm from the lens, its image is formed at (select the correct option)
(A) 20 cm on the other side of the lens and is of the same size, real and erect.
(B) 40 cm on the other side of the lens and is magnified, real and inverted.
(C) 20 cm on the other side of the lens and is of the same size, real and inverted.
(D) 20 cm on the other side of the lens and is of the same size, virtual and erect.
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