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# A convex mirror used as a rear-view mirror in a car has a radius of curvature of 3 m. If a bus is located at a distance of 5 m from this mirror, find the position of image. What is the nature of the image?

Given:

The mirror is convex.

Distance of the object from the mirror, $u$ = $-$5 m

Radius of curvature of the mirror, $R$ = 3 m

Focal length of the mirror, $f$ = 1.5 m $(\because f=\frac{R}{2})$

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

Solution:

From the mirror formula, we know that-

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

Substituting the given values we get-

$\frac {1}{1.5}=\frac {1}{v}+\frac {1}{(-5)}$

$\frac {1}{1.5}=\frac {1}{v}-\frac {1}{5}$

$\frac {1}{5}+\frac {1}{1.5}=\frac {1}{v}$

$\frac {1}{5}+\frac {10}{15}=\frac {1}{v}$

$\frac {1}{v}=\frac {3+10}{15}$

$\frac {1}{v}=\frac {13}{15}$

$v=\frac {15}{13}$

$v=+1.15m$

Thus, the distance of the image $v$ is 1.15 m from the mirror, and the positive sign implies that the image forms behind the mirror (on the right).

Now, from the magnification formula, we know that-

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

Substituting the given values we get-

$m=-\frac {1.15}{(-5)}$

$m=\frac {115}{500}$

$m=+0.22$

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 image is virtual, erect, and small in size.