Question

if an object 4 cm height is placed at a distance of 18 cm from concave mirror having focal length 12 cm.find the position, nature , and height of images​

Answer 1

Given:-

• Object height ,ho = 4cm
• Object distance ,u = -18cm
• Focal length ,f = -12 cm

To Find:-

• Position , nature & height of images.

Solution:-

According to the Question

Firstly we calculate the image distance of the object using mirror formula

• • 1/v + 1/u = 1/f

where,

• v is the Image Position
• u is the object distance
• f is the focal length

Substitute the value we get

$:\implies$ 1/v = 1/f - 1/u

$:\implies$ 1/v = 1/-12 - 1/(-18)

$:\implies$ 1/v = -1/12 + 1/18

$:\implies$ 1/v = -3+2/36

$:\implies$ 1/v = -1/36

$:\implies$ v = -36 cm

• Hence, the image formed is 36 cm from the mirror .

Now, calculating the magnification of the image

• m = hi/ho = -v/u

Substitute the value we get

$:\implies$ hi/4 = -36/-18

$:\implies$ hi/4 = 36/18

$:\implies$ hi/4 = 2

$:\implies$ hi = 8cm

• Hence, the height of image is 8 cm

• Nature of Image :- Real , inverted & larger than the object .

Answer 2

Given :-

If an object 4 cm height is placed at a distance of 18 cm from concave mirror having focal length 12 cm

To Find :-

Position

Nature

Height

Solution :-

1/u + 1/v = 1/f

1/-18 + 1/v = 1/-12

-1/18 + 1/v = -1/12

1/v = -1/12 - (-1)/18

1/v = -3 + 2/36

1/v = -1/36

v = -36 cm

Now

m = -v/u = hi/ho

hi/4 = -(36)/-18

hi/4 = 36/18

hi/4 = 2/1

4 × 2 = hi

8 = hi

For nature - Real, inverted and  larger than the object