Question

Find the position of the object when an image is formed in front of a mirror on a screen at a distance of 30cm from the mirror. The focal length of the mirror is 10cm. Calculate magnification

Answer 1

SOLUTION :

$\underline{\bf{Given\::}}}}$

• Focal length of the mirror, (f) = 10cm
• Distance of object from mirror, (u) = 30 cm
• Distance of image from mirror, (v) = ?

$\underline{\bf{To\:find\::}}}}$

The position & magnification.

$\underline{\bf{Explanation\::}}}}$

Using formula of the Mirror :

$\boxed{\bf{\frac{1}{f} =\frac{1}{v} -\frac{1}{u} }}}}$

$\longrightarrow\sf{\dfrac{1}{-10} =\dfrac{1}{v} -\dfrac{1}{-30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{-10} +\dfrac{1}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{-3+1}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\cancel{\dfrac{-2}{30} }}\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{-15} }\\\\\\\longrightarrow\sf{v=-15\:cm}$

∴ The image is formed at a distance of 15 cm from the mirror same side of object.

$\boxed{\bf{M \:A \:G \:N\: I \:F \:I \:C\: A\: T \:I \:O\: N :}}}$

$\mapsto\sf{m=\dfrac{Height\:of\:image\:(I)}{Height\:of\:object\:(O)} =\dfrac{Distance\:of\:image}{Distance\:of\:object} =\dfrac{v}{u} }\\\\\\\mapsto\sf{m=\cancel{\dfrac{-30}{-15}} }\\\\\\\mapsto\bf{m=2\:cm}$

Thus;

The magnification will be 2 cm .

Answer 2

SOLUTION :

Focal length of the mirror, (f) = 10cm

Distance of object from mirror, (u) = 30 cm

Distance of image from mirror, (v) = ?

The position & magnification.

Using formula of the Mirror :

\begin{lgathered}\longrightarrow\sf{\dfrac{1}{-10} =\dfrac{1}{v} -\dfrac{1}{-30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{-10} +\dfrac{1}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{-3+1}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\cancel{\dfrac{-2}{30} }}\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{-15} }\\\\\\\longrightarrow\sf{v=-15\:cm}\end{lgathered}

⟶

−10

1

=

v

1

−

−30

1

⟶

v

1

=

−10

1

+

30

1

⟶

v

1

=

30

−3+1

⟶

v

1

=

30

−2

⟶

v

1

=

−15

1

⟶v=−15cm

∴ The image is formed at a distance of 15 cm from the mirror same side of object.

\begin{lgathered}\mapsto\sf{m=\dfrac{Height\:of\:image\:(I)}{Height\:of\:object\:(O)} =\dfrac{Distance\:of\:image}{Distance\:of\:object} =\dfrac{v}{u} }\\\\\\\mapsto\sf{m=\cancel{\dfrac{-30}{-15}} }\\\\\\\mapsto\bf{m=2\:cm}\end{lgathered}

↦m=

Heightofobject(O)

Heightofimage(I)

=

Distanceofobject

Distanceofimage

=

u

v

↦m=

−15

−30

↦m=2cm

Thus;

The magnification will be 2 cm .