Question

.An object of height 2 cm is placed at 20 cm distance in front of a concave mirror whose

focal length is 15 cm. Calculate height of the image.​

Answer 1

Answer:

An object of height 2 cm is placed at 20 cm distance in front of a concave mirror whose

focal length is 15 cm. Calculate height of the image.

Answer 2

Height of the image = -6 cm

Given :

➺ Distance of image (u) = - 20 cm

➺ Focal length of the mirror (f) = - 15 cm

➺To Find :

Height of the image (h₂) = ?

To find the height of the image, at first we need to find the distance of image from mirror, after that we will find Height of the image.

We can find distance of the image by using the mirror formula which says:

⊰ 1/v + 1/u = 1/f ⊱

Here,

✎ So let's calculate Distance of image (v) !

$\tt{: \implies \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }:$⟹ v1 + u1 = f1

$\tt{: \implies \dfrac{1}{v} + \dfrac{1}{ - 20} = \dfrac{1}{ - 15} }:$⟹ v1 + −201 = −151

$\tt{: \implies \dfrac{1}{v} - \dfrac{1}{ 20} = - \dfrac{1}{15} }:$⟹ v1 − 201 =− 151

$\tt{: \implies \dfrac{1}{v} = - \dfrac{1}{15} + \dfrac{1}{20}}:$⟹ v1 =− 151 + 201

$\tt{: \implies \dfrac{1}{v} = \dfrac{ - 4 + 3}{60} }:$⟹ v1 = 60−4+3

$\tt{: \implies \dfrac{1}{v} = \dfrac{4 - 3}{60} }:[/tex</p><p>x]⟹ v1 = 604−3</p><p>\tt{: \implies \dfrac{1}{v} = - \dfrac{1}{60} }:⟹ v1 =− 601</p><p> </p><p> </p><p></p><p>[tex]\bf{: \implies \underline{ \: \: \underline{ \orange{ \: \: v = - 60 \: cm\: \: }} \: \: }}:$⟹

v=−60cm

Now, we have Height of the object, distance of the image and distance of the object,

Height of the object (h₁) = 2 cm

Distance of the image (v) = -60 cm

Distance of the object (u) = -20 cm

And we need to find Height of the image (h₂)

We can find Height of the image by using the formula which syas:

⊰ h₂/h₁ = -v/u ⊱

Here,

$\tt{: \implies \dfrac{h_2}{h_1} = - \dfrac{v}{u} }:$⟹ h 1h 2 =− uv

$\tt{: \implies \dfrac{h_2}{2} = - \dfrac{ \cancel{ - }60}{ \cancel{ -} 20}}:$⟹ 2h 2=− −20− 60

$\tt{: \implies \dfrac{h_2}{2} = -\dfrac{6\cancel{0}}{2\cancel{0}}}:$⟹ 2h 2=− 2 06 0

$\tt{: \implies \dfrac{h_2}{2}$ = - \cancel{\dfrac{6}{2}}}:⟹ 2h 2 =− 26

$\tt{: \implies \dfrac{h_2}{2}$ = - 3}:⟹ 2h 2 =−3

$\tt{: \implies h_2 = - 3 \times 2}:$⟹h 2 =−3×2

$\bf{: \implies \underline{ \: \: \underline{ \red{ \: \: h_2 = -6 \: cm \: \: }} \: \: }}:$⟹ h 2 =−6cm

༺Hence, the Height of the image is -6 cm.༻

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