Answer
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Hint: To calculate the image distance and magnification, we are using the mirror image and magnification formula. The mirror image formula gives a relation between the magnification, focal length and the distance of the object.
Formula used:
Mirror image: $\dfrac { 1 }{ f } =\dfrac { 1 }{ v } +\dfrac { 1 }{ u } $
Lateral magnification, $ m=\dfrac { -v }{ u }$
Complete step by step answer:
Focal length is the distance between the center of a convex lens and the focal point of the lens.
The given focal length of the mirror is ‘f’.
Given object distance from the pole, $u=\dfrac { -f }{ 2 }$
The distance between the object and the pole of mirror is called object distance (u)
For a convex lens, the virtual distance is negative and the real distance is positive.
By using mirror formula,
$\dfrac { 1 }{ f } =\dfrac { 1 }{ v } +\dfrac { 1 }{ u }$
We shall substitute the value of ‘u’ in the above formula. On solving, we get
$\dfrac { 1 }{ v } +\dfrac { 1 }{ -\left( \dfrac { f }{ 2 } \right) } =\dfrac { 1 }{ f }$
$\dfrac { 1 }{ v } -\dfrac { 2 }{ f } =\dfrac { 1 }{ f }$
$\dfrac { 1 }{ v } =\dfrac { 1 }{ f } +\dfrac { 2 }{ f }$
$\dfrac { 1 }{ v } =\dfrac { 3 }{ f }$
On reciprocating, it becomes $v=\dfrac { f }{ 3 }$
The distance between the image and the pole of the mirror is called image distance (v).
Using magnification formula and applying the values of ‘u’ and ‘v’, we get
$m=\dfrac { -v }{ u }$
$m=\dfrac { -\left( \dfrac { f }{ 3 } \right) }{ -\left( \dfrac { f }{ 2 } \right) }$
$m=\dfrac { f }{ 3 } \times \dfrac { 2 }{ f }$
$m=\dfrac { 2 }{ 3 }$
Therefore, the image formed is virtual, upright and diminished.
Note: It is important to remember the formulas in optics because only with that, the nature and position of the objects can be determined. It is also to be noted that for a convex mirror, ‘u’ is always negative and ‘f’ is always positive.
Formula used:
Mirror image: $\dfrac { 1 }{ f } =\dfrac { 1 }{ v } +\dfrac { 1 }{ u } $
Lateral magnification, $ m=\dfrac { -v }{ u }$
Complete step by step answer:
Focal length is the distance between the center of a convex lens and the focal point of the lens.
![seo images](https://www.vedantu.com/question-sets/f02ba9dd-eb5a-4de7-bea4-70301c0c2dcd7973966835014265049.png)
The given focal length of the mirror is ‘f’.
Given object distance from the pole, $u=\dfrac { -f }{ 2 }$
The distance between the object and the pole of mirror is called object distance (u)
For a convex lens, the virtual distance is negative and the real distance is positive.
By using mirror formula,
$\dfrac { 1 }{ f } =\dfrac { 1 }{ v } +\dfrac { 1 }{ u }$
We shall substitute the value of ‘u’ in the above formula. On solving, we get
$\dfrac { 1 }{ v } +\dfrac { 1 }{ -\left( \dfrac { f }{ 2 } \right) } =\dfrac { 1 }{ f }$
$\dfrac { 1 }{ v } -\dfrac { 2 }{ f } =\dfrac { 1 }{ f }$
$\dfrac { 1 }{ v } =\dfrac { 1 }{ f } +\dfrac { 2 }{ f }$
$\dfrac { 1 }{ v } =\dfrac { 3 }{ f }$
On reciprocating, it becomes $v=\dfrac { f }{ 3 }$
The distance between the image and the pole of the mirror is called image distance (v).
Using magnification formula and applying the values of ‘u’ and ‘v’, we get
$m=\dfrac { -v }{ u }$
$m=\dfrac { -\left( \dfrac { f }{ 3 } \right) }{ -\left( \dfrac { f }{ 2 } \right) }$
$m=\dfrac { f }{ 3 } \times \dfrac { 2 }{ f }$
$m=\dfrac { 2 }{ 3 }$
Therefore, the image formed is virtual, upright and diminished.
Note: It is important to remember the formulas in optics because only with that, the nature and position of the objects can be determined. It is also to be noted that for a convex mirror, ‘u’ is always negative and ‘f’ is always positive.
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