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# The radius of curvature of a convex mirror used as a rear view mirror in a moving car is 2.0 m. A truck is coming from behind it at a distance of 3.5 m. Calculate** (a)** position, and** (b)** size of the image relative to the size of the truck. What will be the nature of the image?

The mirror is Convex.

Distance of the object, $u$ = $-$3.5 m

Radius of curvature, $R$ = 2 m

Focal length of the mirror, $f$ = 1 m $(f=\frac {R}{2})$

To find: Distance or position of the image, $v$, the magnification, $m$, and the nature of the image.

Solution:

**(a)** From the mirror formula, we know that-

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

Substituting the given values we get-

$\frac {1}{1}=\frac {1}{v}+\frac {1}{(-3.5)}$

$\frac {1}{1}=\frac {1}{v}-\frac {1}{3.5}$

$1+\frac {1}{3.5}=\frac {1}{v}$

$\frac {1}{v}=\frac {3.5+1}{3.5}$

$\frac {1}{v}=\frac {4.5}{3.5}$

$\frac {1}{v}=\frac {45}{35}$

$\frac {1}{v}=\frac {9}{7}$

$v=\frac {7}{9}$

$v=+0.77m$

Thus, the distance of the image $v$ is 0.77 m from the mirror, and the positive sign implies that the image forms behind the mirror (on the right).

**(b) **From the magnification formula, we know that-

$m=-\frac {v}{u}$

Substituting the given values we get-

$m=-\frac {0.77}{(-3.5)}$

$m=\frac {77}{350}$

$m=+0.22$

Thus, the magnification is $0.22$ which is less than 1, which means the image is small in size, and the positive sign implies that the image is virtual and erect.

Hence, the nature of the image is virtual, erect, and small in size.