
The magnification is more than unity when the object is placed at a distance $d$ from a convex lens. Its focal length is $20cm$. What is the value of $d$?
(a) Greater than $20cm$
(b) Less than $20cm$
(c) $40cm$
(d) Greater than $20cm$and less than $40cm$
Answer
473.4k+ views
Hint:The magnification of a lens is defined as the ratio of image distance to that of object distance. We know that the value of magnification given in this question is greater than unity. In this equation, we will replace the value of magnification with the ratio of image distance to that of object distance and get the value of $d$.
Complete step by step answer:
In this question we are given that the magnification is greater than unity, so,
$m > 1......(1)$
We also know that the magnification is the ratio of image distance to that of object distance. So,
$m = \dfrac{v}{u}$
On putting the above value in the equation (1), we get,
$\dfrac{v}{u} > 1$
This can further be written as,
$v > u$
This means that for magnification to be greater than unity, the image distance must be greater than the object distance. This is only possible when the object is placed at a distance less than the focal length of the convex lens.
In this question we are given that the focal length of the convex lens is $20cm$, so the object is placed at a distance less than $20cm$.
Hence, the value of $d$ is less than $20cm$.
Note:When the object is placed at a distance less than the focal length in front of a convex lens, i.e., between the focus and the pole, then a virtual image is formed. The size of this image is larger than that of the object and hence, the magnification is greater than unity.
Complete step by step answer:
In this question we are given that the magnification is greater than unity, so,
$m > 1......(1)$
We also know that the magnification is the ratio of image distance to that of object distance. So,
$m = \dfrac{v}{u}$
On putting the above value in the equation (1), we get,
$\dfrac{v}{u} > 1$
This can further be written as,
$v > u$
This means that for magnification to be greater than unity, the image distance must be greater than the object distance. This is only possible when the object is placed at a distance less than the focal length of the convex lens.
In this question we are given that the focal length of the convex lens is $20cm$, so the object is placed at a distance less than $20cm$.
Hence, the value of $d$ is less than $20cm$.
Note:When the object is placed at a distance less than the focal length in front of a convex lens, i.e., between the focus and the pole, then a virtual image is formed. The size of this image is larger than that of the object and hence, the magnification is greater than unity.
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