Question

a convex mirror has a focal length of 18cm the image of an object kept in front of the mirror is half the height of the object what is the distance of the object from the mirror

Answer 1

Let the Height of the object be = '2h'

So the height of the image formed would be ='h'

Focal Length of convex mirror is always positive, so f= '18 cm'

magnification= Height of image/Height of object = -V(image dis.)/U(object dis.)

Magnification = h/2h = 1/2

1/2 = -V/U

-2V=U

USING MIRROR FORMULA-: 1/f = 1/U +1/V

1/18 = 1/V + 1/-2V

1/18 = 1/V -1/2V

1/18 = 2/2V - 1/2V

1/18 = 1/2V

2V = 18

V = 9,

1/18 - 1/9 = 1/U

-1/18= 1/U

U = -18

So In my opinion, the final answer should be = -18,  you may cross check it once because my claculations are very weak.

If you find the answer to be helpful and correct, please mark my answer as Brainliest.

Thanks.....

Answer 2

The distance of the object from the mirror is 54 cm.

Explanation:

Given :

A convex mirror has focal length of 18 cm. The image of an object kept in front of the mirror is half of the height of the object.

To find : What is the distance of the object from the mirror?

Solution :

A convex mirror has focal length of 18 cm.

Let the object placed at O=x cm.

The image of an object kept in front of the mirror is half of the height of the object.

So, Image is at $I=\frac{x}{2}cm$

The formula to find the relationship is

$\frac{I}{O}= \frac{v}{u}$

$\frac{\frac{x}{2}}{x}= \frac{v}{u}$

$v= \frac{u}{2}$

Again the focal length formula is

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

Substitute the values,

$\frac{1}{18}= \frac{1}{u}+\frac{1}{\frac{u}{2}}$

$\frac{1}{18}= \frac{1}{u}+\frac{2}{u}$

$\frac{1}{18}= \frac{3}{u}$

Cross multiply,

$u=3\times 18$

$u=54$

Therefore, The distance of the object from the mirror is 54 cm.✅

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