
Under what condition will the magnification produced by a thin lens be -1?
Answer
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Hint: Magnification is defined as a ratio of the sizes of the image and the object. Mathematically it is expressed as $m = \dfrac{{{h'}}}{h}$ where m is the magnification of the image, ${h'}$ is the size of the image and h is the size of the object.
It tells us the factor by which an image is scaled with respect to the original object.
There is one more way of expressing the magnification produced by a lens. Magnification is also described as the ratio of the image distance with the object distance.
Mathematically this relation is expressed as $m = \dfrac{v}{u}$ where m is the magnification of the image, v is the image distance and u is the object distance.
Complete step by step answer:
Magnification is described in two terms.
In terms of the distance of the object and its image
In terms of the size of the object and its image
In terms of the distance of the object and its image,
Mathematically it is expressed as $m = \dfrac{v}{u}$ where m is the magnification of the image, v is the image distance and u is the object distance.
Given that $m = - 1$ ,
Hence, $m = \dfrac{v}{u} = - 1$
$ \Rightarrow v = - u$
This implies that the image distance is equal to the object distance in magnitude. However, the image is formed to the right of the lens while the object is on the left side of the lens.
In terms of the size of the object and its image,
Mathematically it is expressed as $m = \dfrac{{{h'}}}{h}$ where m is the magnification of the image, ${h'}$ is the size of the image and h is the size of the object.
Given that $m = - 1$ ,
Hence, $m = \dfrac{{{h'}}}{h} = - 1$
$ \Rightarrow {h'} = - h$
This implies that the image size is equal to the object distance in magnitude. However, the image is inverted.
Note:
While dealing with lenses, keep a proper note of the conventions to be followed. We always take the light to enter the lens through the left side of the lens and the object is also kept towards the left. The negative sign of the magnification shows that the image formed is real and inverted. When magnification is less than 1 we say that the image is diminished while when it is greater than 1 we say that the image is enlarged.
It tells us the factor by which an image is scaled with respect to the original object.
There is one more way of expressing the magnification produced by a lens. Magnification is also described as the ratio of the image distance with the object distance.
Mathematically this relation is expressed as $m = \dfrac{v}{u}$ where m is the magnification of the image, v is the image distance and u is the object distance.
Complete step by step answer:
Magnification is described in two terms.
In terms of the distance of the object and its image
In terms of the size of the object and its image
In terms of the distance of the object and its image,
Mathematically it is expressed as $m = \dfrac{v}{u}$ where m is the magnification of the image, v is the image distance and u is the object distance.
Given that $m = - 1$ ,
Hence, $m = \dfrac{v}{u} = - 1$
$ \Rightarrow v = - u$
This implies that the image distance is equal to the object distance in magnitude. However, the image is formed to the right of the lens while the object is on the left side of the lens.
In terms of the size of the object and its image,
Mathematically it is expressed as $m = \dfrac{{{h'}}}{h}$ where m is the magnification of the image, ${h'}$ is the size of the image and h is the size of the object.
Given that $m = - 1$ ,
Hence, $m = \dfrac{{{h'}}}{h} = - 1$
$ \Rightarrow {h'} = - h$
This implies that the image size is equal to the object distance in magnitude. However, the image is inverted.
Note:
While dealing with lenses, keep a proper note of the conventions to be followed. We always take the light to enter the lens through the left side of the lens and the object is also kept towards the left. The negative sign of the magnification shows that the image formed is real and inverted. When magnification is less than 1 we say that the image is diminished while when it is greater than 1 we say that the image is enlarged.
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