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# A large concave mirror has a radius of curvature of 1.5 m. A person stands 10 m in front of the mirror. Where is the person's image?

Given:

Radius of curvature, $R$ = $-$1.5 cm

Distance of the object, $(u)$ = $-$10 cm

To find: Distance of the image $(v)$ from the mirror.

Solution:

We know that,

$f=\frac {R}{2}$, where, $f$ = focal length, and $R$ = radius of curvature.

Putting the vlaue of $R$, we get-

$f=\frac {-1.5}{2}$

$f=-0.75cm$

So, the focal length is **0.75 c**m.

Now, from the mirror formula, we know that-

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

Substituting the given values in the mirror formula we get-

$\frac{1}{(-0.75)}=\frac{1}{v}+\frac{1}{(-10)}$

$-\frac{1}{0.75}=\frac{1}{v}-\frac{1}{10}$

$\frac{1}{10}-\frac{100}{75}=\frac{1}{v}$

$\frac{1}{v}=\frac{15-200}{150}$

$\frac{1}{v}=\frac{-185}{150}$

$\frac{1}{v}=\frac{-37}{30}$

$v=-\frac{30}{37}$

$v=-0.81m$

Thus, the distance of the image, $v$ is 0.81 m.

Hence, the person's image will be formed at a distance of **0.81 m** from the mirror.

And, the negative sign implies that the image is formed in front of the mirror (on the left).