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# An object is placed 15 cm from **(a) **a converging mirror, and **(b)** a perging mirror, of radius of curvature 20 cm. Calculate the image position and magnification in each case.

**(a) It is a converging mirror, which means concave mirror.**

Distance of the object from the mirror, $u$ = $-$15 cm

Radius of curvature, $R$ = $-$20 cm

Then, the focal length of the mirror, $f$ = $-$10 cm $[\because f=\frac {1}{2}\times {R}]$

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

Solution:

From the mirror formula, we know that-

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

Substituting the given values we get-

$\frac {1}{(-10)}=\frac {1}{(-15)}+\frac {1}{v}$

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

$\frac {1}{15}-\frac {1}{10}=\frac {1}{v}$

$\frac {1}{v}=\frac {2-3}{30}$

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

$v=-30cm$

Thus, the distance of the image $v$ is **30 cm** from the mirror, and the negative sign implies that the image forms in front of the mirror (on the left).

Now, from the magnification formula, we know that-

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

Substituting the given values we get-

$m=-\frac {(-30)}{(-15)}$

$m=\frac {-30}{15}$

$m=-2$

Thus, the magnification is $2$ which is more than 1, which means the image is large in size, and the negative sign implies that the image is real and inverted.

Hence, the image is real, inverted, and large in size.

**(b) It is a diverging mirror, which means convex mirror.**

**Given:**

Distance of the object from the mirror, $u$ = $-$15 cm

Radius of curvature, $R$ = 20 cm

Then, the focal length of the mirror, $f$ = 10 cm $[\because f=\frac {1}{2}\times {R}]$

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

**Solution:**

From the mirror formula, we know that-

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

Substituting the given values we get-

$\frac {1}{10}=\frac {1}{(-15)}+\frac {1}{v}$

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

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

$\frac {1}{v}=\frac {2+3}{30}$

$\frac {1}{v}=\frac {5}{30}$

$v=+6cm$

Thus, the distance of the image $v$ is **6 cm **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-

$m=-\frac {6}{(-15)}$

$m=\frac {2}{5}$

$m=+0.4$

Thus, the magnification is $0.4$ 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.