Question

An object is kept at a distance of 15 cm from a concave mirror whose radius of curvature is 30 cm. What is the position of the images?

Answer 1

i know this answer

first we do 30+15 =45
so the mirror image is 45 cm

Answer 2

Answer:

The position of the image is 15cm, i.e., at the focus.

Explanation:

Given:

object distance (u) = 15cm

radius of curvature (r) = 30cm

To find:

image distance (v)

Formula:

$f=\frac{r}{2}$

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

Step 1:

To find the focal length of the concave mirror, we divide the radius of curvature by 2.

$f=\frac{r}{2}$

$f=\frac{30}{2}$

$f=15cm$

Step 2:

From the sign conventions of a concave mirror, object distance and focal length are taken to be negative as they lie on the left-hand side of the mirror.

Therefore, $u=-15cm$ and $f=-15cm$

Step 3:

Substituting the given values.

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

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

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

$\frac{1}{v} = \frac{(-15)-(-15)}{225}$

$v=0$

This implies that the image is formed at the same position as the object and the distance between them is minimum.

Therefore, when the object is placed at the focus, the image is also obtained at the focus.

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