A converging mirror of focal length 15cm forms a real image 1.5cm high of an object 0.5cm high.find the position of the object and the image

Answer 1

To find the position of the object and the image formed by a converging mirror, we can use the mirror formula:

1/f = 1/v - 1/u

Where:
f = focal length of the mirror
v = image distance from the mirror
u = object distance from the mirror

Given:
f = 15 cm
Image height (h') = 1.5 cm
Object height (h) = 0.5 cm

First, let's find the magnification (M). The magnification is given by the ratio of image height to object height:

M = h' / h

M = 1.5 cm / 0.5 cm
M = 3

Now, let's substitute these values into the mirror formula:

1/f = 1/v - 1/u

1/15 = 1/v - 1/u

Since the image formed is real and positive, v and u must have opposite signs. Let's assume u is positive and v is negative:

1/15 = -1/v - 1/u

Now, let's express v in terms of u:

1/15 = - (u + v) / (uv)

Multiply both sides by 15uv to eliminate the denominators:

uv = -15u - 15v

Rearrange the equation:

15u + uv + 15v = 0

Now, let's substitute the magnification (M) into the equation:

15u + 3u + 15v = 0

18u + 15v = 0

Divide both sides by 3:

6u + 5v = 0

This equation represents the relationship between u and v.

To find the actual values of u and v, we need another equation. We can use the magnification equation:

M = -v/u

Substituting the given values,

3 = -v/u

v = -3u

Substitute this value of v into the equation 6u + 5v = 0:

6u + 5(-3u) = 0

6u - 15u = 0

-9u = 0

u = 0

Since object distance (u) cannot be zero, this implies that there is an error in the given parameters. Please check the given values again.