Question

A concave mirror used on an automobile has a radius of curvature 3.0 metre. If a truck is following it at a constant distance of 4.5 metre find the position, the nature and magnification for the image ?

Answer 1

$\bf{\underline{\red{Correct\:Question:-}}}$

✓✓A convex mirror used on a moving automobile has a radius of curvature of 3.0 m. If a truck is following it at a constant distance of 4.5 m, find.

• (i) the position,
• (ii) the nature and
• (iii) the magnification for the image.

$\bf{\underline{\blue{Given:-}}}$

• $\bf\:Radius\:of\:curvature,\:R = 3.0m$  

⠀⠀⠀⠀ ⠀ ⠀⠀[R is +ve for a convex mirror]

• $\bf\:Object\:distance,\:u = - 4.5 m$

• $\bf\:Image\:distance,\:v = ?$

• $\bf\:Image\:size,\:h' = ?$

$\bf{\underline{\purple{Solution:-}}}$

According to focal length and radius of curvature ,we have

$\bf\:Focal\: length\:(f)=\dfrac{R}{2}$

$\bf\implies\:f=\dfrac{3.0m}{2}$

$\bf\implies\:f=1.50\:m$

[ f is positive for a convex mirror]

$\bf\boxed\star\red{\underline{\underline{{Using\:the\:mirror\: formula:-}}}}$

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

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

$\bf\implies\dfrac{1}{v}=\dfrac{1}{1.50}-\dfrac{1}{-4.5}$

$\bf\implies\dfrac{1}{v}=\dfrac{1}{1.50}+\dfrac{1}{4.5}$

$\bf\implies\dfrac{1}{v}=\dfrac{3+1}{4.5}$

$\bf\implies\dfrac{1}{v}=\dfrac{4}{4.5}$

$\bf\implies\:v=1.125$

$\bf\green{{\:Image\: distance\:(v)=1.125}}$

$\bf\: Magnification\:(m)=\dfrac{h'}{h}=-\dfrac{v}{u}$

$\bf\implies\:m=\dfrac{1.125}{4.5}$

$\bf\implies\red{{m=0.25}}$

$\bf\therefore$The image is formed at a distance of 1.125 m behind the mirror.

The image formed is virtual, erect and smaller in size .