Question

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a concave mirror has a focal length of 20 cm . at what distance from the mirror should a 4 cm tall object be placed so that it forms an image at a distance of 30 cm from the mirror . also calculate the size of image
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Answer 1

Answer: The height of the image is 2 cm and it is real and inverted. The object is placed at 60 cm from the mirror.

Explanation:

Given that,

Focal length f = -20 cm

Height of the object h = 4 cm

Distance of the image v = -30 cm

Using mirror's formula

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

$\dfrac{1}{-20}=\dfrac{1}{-30}+\dfrac{1}{u}$

$\dfrac{1}{u}=- \dfrac{1}{60}$

$u = -60 cm$

The object is placed at 60 cm from the concave mirror.

The magnification is

$m = \dfrac{h'}{h}=-\dfrac{v}{u}$

$\dfrac{h'}{4}=-\dfrac{-30}{-60}$

$h' = -2 cm$

Hence, The height of the image is 2 cm and it is real and inverted. The object is placed at 60 cm from the mirror.

Answer 2

Given:

Focal length,

f = -20 cm

Object's height,

h = 4 cm

Distance of the image,

v = -30 cm

To Find:

Size of the image = ?

Solution:

On applying Mirror's formula, we get

⇒  $\frac{1}{f} =\frac{1}{v} +\frac{1}{u}$

On putting the estimated values, we get

⇒  $\frac{1}{-20} =\frac{1}{-30} +\frac{1}{u}$

⇒  $\frac{1}{u} =-\frac{1}{20} +\frac{1}{30}$

⇒  $\frac{1}{u} =\frac{-3+2}{60}$

⇒  $\frac{1}{u} =-\frac{1}{60}$

⇒  $u=-60 \ cm$

Now,

The magnitude will be:

⇒  $m=\frac{{h}'}{h} =-\frac{v}{u}$

On putting the values, we get

⇒  $\frac{{h}'}{4} =\frac{-30}{-60}$

On applying cross-multiplication, we get

⇒  $-60\times {h}'=-30\times 4$

⇒  $-60\times {h}'=-120$

⇒             ${h}'=\frac{-120}{-60}$

⇒             ${h}'=-2 \ cm$

Therefore, image's height will be "-2 cm".