Question

Find the position nature and size of the image of an object 3 cm high placed at a distance of 9 cm from a concave mirror of focal length 18 cm (v = 18cm , h = 6cm)

Answer 1

Hey !! ^_^

Here is your answer

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Size of object(o) = 3cm

Distance of object (u) = -9cm ( sign convention)

focal length(f) = -18cm ( sign convention)


Mirror Formula

1/f = 1/v + 1/u

1/v = 1/f - 1/u

putting the value


1/v = 1/(-18) - 1/(-9)

1/v = 1/(-18) + 1/9

1/v = (-1 + 2)/18

1/v = 1/18

v = 18.( + sign indicate that the image is formed behind the mirror)

The position of image is behind the mirror 18 cm far from the pole

m = v/u

m = 18/9

m = 2

The size of image is 2 times larger than object or magnified

or ...

m = I/O

2 = I/3

I = 3 × 2

I = 6


Image will be virtual and erect


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Answer 2

Answer:

The image will formed at 18 cm behind the mirror. The image is virtual and erect.

Explanation:

Given that :

• Height of the object = 3 cm
• Distance of object from pole, u = - 9 cm
• Focal length of concave mirror = - 18 cm

Solution :

• The mirror formula is a mathematical relationship between object distance u, image distance v, and focal length f of a spherical mirror.
• Thia relation is

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

• Here, u = -9 cm, f = -18 cm

$- \frac{1}{18} = \frac{1}{v} + \frac{1}{ - 9} \\ \frac{1}{v} = \frac{1}{9} - \frac{1}{18} \\ \frac{1}{v} = \frac{1}{18} \\ v = 18 \: cm$

• v = 18 cm.
• Image will formed behind the mirror.
• The ratio of height of the image to that of the object is called linear magnification.

$\frac{H}{h} = - \frac{v}{u} \\ H = - \frac{18}{ - 9} \times 3 \: cm \\ H = 6 \: cm$

• Magnification is 2. Height of the image is 2 times that of the object.
• Height of the image will be 6 cm.
• The image formed is virtual and erect.