Question

an object 5cm in length is placed at a distance of 20cm in front a convex mirror of radius of curvature 30cm . Find the curvature
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Answer 1

Answer:

Size of the object O = 5cm, u =20cm, f=r/2=30/2=15cm

hence by mirror formula,

1/f=1/v+1/u

1/15=1/v+1/20

1/v=1/15 -1/20

hence, v=60 cm

it means a virtual and erect image is formed behind the mirror and size of the image,

m= -V/U=I/O

I =( -V/U)× O

= -60×5/20

=-15 CM

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Answer 2

Appropriate Question:

An object 5cm in length is placed at a distance of 20cm in front a convex mirror of radius of curvature 30cm . Find the nature of the image formed .

Solution :

It is given that :

Height of the object = 5 cm

Distance of Object = ( -20 ) cm

Radius of Curvature= 30 cm

Focal length = 30/2 = 15 cm

By using Mirror Formula :

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rightarrow\boxed { \sf{ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} }}$

Here, Image Distance is " v " and Object Distance is " u ". Therefore By Substituting the values ;

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: { \sf{ \dfrac{1}{15} = \dfrac{1}{v} + \dfrac{1}{( - 20)} }} \\ \\ { \sf{ \dfrac{1}{15} = \dfrac{1}{v} - \dfrac{1}{\\ 20} }} \\ \\ \sf{} \frac{1}{v} = ( - \frac{1}{15} ) - \frac{1}{20} \\ \\ \sf{} \frac{1}{v} = \frac{ - 4 - 3}{60} \\ \\ \sf{} \frac{1}{v} = \frac{ - 7}{60} \\ \\ \sf{} v = ( - \frac{60}{7}) \\ \\ \boxed{\sf{}v = - 8.5 \: cm \: }$

Therefore, Image Distance is equals to - 8.5 cm

Now to Find the nature of the image formed we have to calculate magnification. Which is calculated by ;

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rightarrow\boxed { \sf{m = \frac{-v}{u} }}$

By putting Image Distance and Object Distance ;

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: { \sf{m = \dfrac{-(-8.5)}{ - 20} }} \\ \\ { \sf{ m = \cancel{\dfrac{ 8.5}{ - 20} }}} \\ \\ \boxed {\sf{ m = - 0.42}}$

From th value of m < 1 , we can conclude that the image formed Is Diminished in size. which means Size of Object is smaller than size of Image .

Now , from that negative sign of image we can conclude that the image formed is Erect and Virtual .

Now , height of the image can be calculated by :

$\boxed{\sf{m = \dfrac{Height_{(image)}}{Height_{(object)}}}}$

By substututing the values we get,

${\sf{-0.425 = \dfrac{Height_{(image)}}{5}}}$

${\sf{Height_{(image)}= 5 × ( - 0 .425) }}$

${\sf{Height_{(image)}= - 2.14\: cm }}$

Therefore, height of the image formed is equals to -2.14 cm.