Question

Question ⤵
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4.5 cm Niddle is placed 12 cm away from a convex mirror of focal Lenght 1 cm.
Give the location of image and Magnification?
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Answer 1

$\mathbb{ANSWER \: WITH \: CLEAR \: SOLUTION}$

Given: Height of the needle, $H_{1}$​= 4.5cm.

Object distance, u= −12cm.

The focal length of the convex mirror, f=15cm.

Image distance, v The value of v can be obtained using the mirror formula.

v/1 ​+ u/1  ​ =  f/1​

1/v + 1/-12 = 1/15

v/1​= 12/1 ​+ 151​

v1​=609​

∴ v  ≈  6.7cm

Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror.

The image size is given by the magnification formula.

m =  h′​/h = −v/u

h′ = $\frac{6.7*4.5}{12}$

⇒ h′ =  +2.5cm

So, m =  $\frac{2.5}{4.5}$

    m   =   0.56

The height of the image is 2.5cm. The positive sign indicates that the image is erect, virtual, and diminished. If the needle is moved farther from the mirror, the size of the image will reduce gradually.

$\rightarrow\mathcal{HOPE \: IT\: HELPS}\leftarrow$

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Answer 2

$\Large{\underbrace{\underline{\sf{Understanding\; the\; Question}}}}$

Here in this question, we have given that a   needle  is placed 12cm away from convex mirror of focal length 1cm. We are asked to find location of image and the magnification. To find location of image and magnification of image we have to apply mirror formula to find image distance.

So let's do it!!

Given:-

• Distance of needle, u=-12 cm
• Focal length=1 cm
• Height of needle, h=4.5cm

Solution:-

We have mirror formula:

$\underline{\large{\boxed{\sf\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}}}}\bigstar$

$\sf \longmapsto \dfrac{1}{1}=\dfrac{1}{v}+\dfrac{1}{-12}$

$\sf \longmapsto \dfrac{1}{1}=\dfrac{1}{v}-\dfrac{1}{12}$

$\sf \longmapsto \dfrac{1}{1}+\dfrac{1}{12}cm=\dfrac{1}{v}$

$\sf \longmapsto \dfrac{12+1}{12}cm=\dfrac{1}{v}$

$\sf \longmapsto \dfrac{13}{12}cm=\dfrac{1}{v}$

$\sf \longmapsto \dfrac{12}{13}cm=v$

$\underline{\boxed{\sf{\red{0.99cm=v}}}}\star$

If means image will form 0.99cm from the mirror. And the position of the mirror is between focus and pole.

Now we have formula for magnification:

$\underline{\boxed{\sf Magnification,m=\dfrac{-v}{u}}}\star$

$\sf \longmapsto Magnification,m=\dfrac{-0.99}{12}$

$\sf \longmapsto Magnification,m=0.09\; approx$

Since magnification is less than 1, the image formed will be highly diminished.

Required Answer :-

• Image will form 0.99cm from the mirror, and the position of the mirror is between focus and pole.

• Magnification = 0.09