Question

convex mirror used on a moving automobile has a radius of curvature 4 m. If a truck is

following it at a constant distance of 6 m, find the position, nature and magnification of the

image.​

Answer 1

Answer :-

→ Position of the object is 1.5 m from the mirror.

→ Nature of the image is erect and virtual .

→ Magnification is +0.25 .

Explanation :-

We have :-

• Radius of curvature = 4 m

• Distance of the object = 6 m

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Focal length of the mirror :-

⇒ f = R/2

⇒ f = 4/2

⇒ f = 2m

Since the mirror is convex :-

• u = -6 m

• f = + 2 m

According to mirror formula :-

1/v + 1/u = 1/f

⇒ 1/v = 1/f - 1/u

⇒ 1/v = 1/2 - 1/(-6)

⇒ 1/v = 1/2 - (-1)/6

⇒ 1/v = (3 + 1)/6

⇒ 1/v = 4/6

⇒ 6 = 4v

⇒ v = 4/6

⇒ v = 1.5 m

[As 'v' is +ve, image is formed on the right of the mirror].

______________________________

Now, from formula of magnification :-

m = -(v/u)

⇒ m = -(1.5/-6)

⇒ m = + 0.25

Answer 2

Appropriate Question:-

A Convex mirror used on a moving automobile has a radius of curvature 4 m. If a truck is following it at a constant distance of 6 m, find the position, nature and magnification of the image.

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Given:-

• Radius of curvature = 4 m

• Distance of object = 6 m

To Find:-

• Position, Nature and Magnification of the image.

Formula Used:-

• ${\boxed{\bf{f=\dfrac{R}{2}}}}$

• ${\boxed{\bf{Mirror\: Formula:\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}}}}$

• ${\boxed{\bf{Magnification:m=\dfrac{-v}{u}}}}$

Solution:-

Firstly,

$\tt :\implies\:f=\dfrac{R}{2}$

$\tt :\implies\:f=\dfrac{4}{2}$

$\sf :\implies\:f=2\:m$

Now, Using

$\bf :\implies\:\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

Here,

• u = -6 m

• f = 2 m

Putting Values,

$\tt :\implies\:\dfrac{1}{2}=\dfrac{1}{v}+\dfrac{1}{-6}$

$\tt :\implies\:\dfrac{1}{2}=\dfrac{1}{v}-\dfrac{1}{6}$

$\tt :\implies\:\dfrac{1}{v}=\dfrac{1}{2}+\dfrac{1}{6}$

$\tt :\implies\:\dfrac{1}{v}=\dfrac{3+1}{6}$

$\tt :\implies\:\dfrac{1}{v}=\dfrac{4}{6}$

$\tt :\implies\:\dfrac{1}{v}=\dfrac{2}{3}$

$\tt :\implies\:v=\dfrac{3}{2}$

$\bf :\implies\:v=1.5\:m$

Hence, image is 1.5m away from the mirror on the right side as v is positive.

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Now,

$\tt :\implies\:m=\dfrac{-v}{u}$

$\tt :\implies\:m=\dfrac{-(1.5)}{-6}$

$\tt :\implies\:m=\dfrac{15}{60}$

$\tt :\implies\:m=\dfrac{1}{4}$

$\bf :\implies\:m=0.25$

Hence, The Magnification is +0.25 and the image is magnified.

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