Question

the radius of curvature of a convex mirror used in an automobile is 200 CM a bus is Coming from behind it at a distance of 3.5 metre calculate the position and size of image relative to but also comment on the nature of image calculate all calculation in metre​.

Answer 1

★Given :

• The radius of curvature of a convex mirror used in an automobile is 200 cm.

• A bus is coming from behind it at a distance of 3.5 metre.

★To Find :

• Position of image.

• Relative size of image(Magnification)

• Nature of image.

★Solution :

As the given mirror is convex,the focal length is positive.

The object distance is negative.

We know :

$\leadsto \boxed{\sf f = \frac{R}{2} }$

Where,

• f = focal length = 200cm
• R = radius of curvature

Therefore,

→f = 200/2

→f = 100 cm

→f = 1 m  

[As 100cm=1m]

~To find the position of image :

Using the mirror formula,

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

Substituting the values,

$\implies \sf \dfrac{1}{v} - \dfrac{1}{3.5 } = \dfrac{1}{1}$

$\implies \sf \dfrac{1}{v} = 1 + \dfrac{1}{3.5}$

$\implies \sf \dfrac{1}{v} = \dfrac{3.5 + 1}{3.5}$

$\implies \sf \dfrac{1}{v} = \dfrac{4.5}{3.5}$

$\implies \sf v = \dfrac{3.5}{4.5}$

$\implies \boxed{\sf v = 0.77}$

∴ Position of image is 0.77 m.

~To Find relative size of the image :

Using the formula,

$\leadsto \underline{\boxed{\sf M = \dfrac{-v}{u} }}$

Substituting values,

$\implies \sf m = \dfrac{\not{-}0.77}{\not{-}3.5}$

$\implies \boxed{\sf M = 0.22}$

∴ Relative to the size of object,the size of image is 0.22 times of object.

~To find nature of image :

We have the position of image is +0.77m.Hence,the image is formued behind the mirror which means it is virtual and erect.

We have got the magnification as 0.22.

In case of virtual and errect image, size of image and object both are positive.So the magnification should be positive.

Hence if the magnification is positive, the image is upright compared to the object; if m is negative, the image is inverted compared to the object.So we can say here the image is diminished.

Therefore,

The nature of image is virtual erect and diminished.

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Answer 2

★Given :

The radius of curvature of a convex mirror used in an automobile is 200 cm.

A bus is coming from behind it at a distance of 3.5 metre.

★To Find :

Position of image.

Relative size of image(Magnification)

Nature of image.

★Solution :

As the given mirror is convex,the focal length is positive.

The object distance is negative.

We know :

$\leadsto \boxed{\sf f = \frac{R}{2} }$

Where,

f = focal length = 200cm

R = radius of curvature

Therefore,

→f = 200/2

→f = 100 cm

→f = 1 m  

[As 100cm=1m]

~To find the position of image :

Using the mirror formula,

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

Substituting the values,

$\implies \sf \dfrac{1}{v} - \dfrac{1}{3.5 } = \dfrac{1}{1}$

$\implies \sf \dfrac{1}{v} = 1 + \dfrac{1}{3.5}$

$\implies \sf \dfrac{1}{v} = \dfrac{3.5 + 1}{3.5}$

$\implies \sf \dfrac{1}{v} = \dfrac{4.5}{3.5}$

$\implies \sf v = \dfrac{3.5}{4.5}$

$\implies \boxed{\sf v = 0.77}$

∴ Position of image is 0.77 m.

~To Find relative size of the image :

Using the formula,

$\leadsto \underline{\boxed{\sf M = \dfrac{-v}{u} }}$

Substituting values,

$\implies \sf m = \dfrac{\not{-}0.77}{\not{-}3.5}$

$\implies \boxed{\sf M = 0.22}$

∴ Relative to the size of object,the size of image is 0.22 times of object.

~To find nature of image :

We have the position of image is +0.77m.Hence,the image is formued behind the mirror which means it is virtual and erect.

We have got the magnification as 0.22.

In case of virtual and errect image, size of image and object both are positive.So the magnification should be positive.

Hence if the magnification is positive, the image is upright compared to the object; if m is negative, the image is inverted compared to the object.So we can say here the image is diminished.

Therefore,

The nature of image is virtual erect and diminished.

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