Question

an object is placed at a distance of 9cm from convex mirror of focal length 15cm.find position, nature and magnification of the image​

Answer 1

Answer:

When an object is placed at a distance of 15 cm from a concave mirror its image is formed at 10 cm in front of the mirror. When an object is placed at a distance of 15 cm from a concave mirror, its image is formed at 10 cm in front of the mirror.

Explanation:

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Answer 2

Answer:

The distance of the image is 5.6cm,  the linear magnification is 0.6, and the nature of the image is virtual and erect.

Explanation:

Given  that, the distance of the object (u) = -9cm

Focal length of the mirror (f) = 15cm

Now, as we know that the mirror formula

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

So, we  put all the given values in the mirror formula.

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

⇒ $\frac{1}{-9} + \frac{1}{v} = \frac{1}{15}$

⇒ $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

⇒ $\frac{1}{v} = \frac{1}{15} - \frac{1}{-9}$ = $\frac{1}{v} = \frac{1}{15} + \frac{1}{9}$

⇒ $\frac{1}{v} = \frac{3 +5}{45}$ = $\frac{8}{45}$

⇒ v = $\frac{45}{8}$ = 5.6cm

Now, as we know that the formula of linear magnification.

⇒  m = $-\frac{v}{u}$ = $-\frac{5.6}{-9} = 0.6$

Therefore, here we can see that the positive sign shows that the image is virtual and erect.

Hence, the distance of the image is 5.6cm,  the linear magnification is 0.6, and the nature of the image is virtual and erect.

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