Solution:

Given data:
Distance of the image from the optical center of the lens = 25cm
Distance of the object from the optical center of the lens = ?
Magnification of the image = -1

Formula used:
Magnification (m) = -v/u
where v is the distance of the image from the optical center of the lens, and u is the distance of the object from the optical center of the lens.

Calculation:
Magnification (m) = -1
Distance of the image from the optical center of the lens (v) = 25cm
Distance of the object from the optical center of the lens (u) = ?

Using the formula, we get:
-1 = -25/u
u = 25cm

Therefore, the object is placed at a distance of 25cm from the optical center of the lens.

To find the natural and focal length of the lens, we use the formula:
f = v u / (v + u)
where f is the focal length of the lens.

Using the given data, we get:
f = 25 x 25 / (25 + 25) = 12.5cm

Therefore, the natural length of the lens is 12.5cm.

To find the focal length of the lens, we use the formula:
1/f = 1/v + 1/u

Substituting the values, we get:
1/f = 1/25 + 1/25
1/f = 2/25
f = 25/2
f = 12.5cm

Therefore, the focal length of the lens is also 12.5cm.

If the object is displaced 15cm towards the optical center of the lens, the new distance of the object from the optical center of the lens (u') is given by:
u' = u - 15
u' = 25 - 15
u' = 10cm

To find the new distance of the image from the lens (v'), we use the formula:
1/f = 1/v' + 1/u'

Substituting the values, we get:
1/12.5 = 1/v' + 1/10
Solving for v', we get:
v' = -20cm

Therefore, the image is formed 20cm to the left of the lens.

Ray diagram:
To justify the answer, we can draw a ray diagram using the following steps:
1. Draw a horizontal line representing the principal axis of the lens.
2. Draw a vertical line representing the object, and mark it at a distance of 25cm from the lens.
3. Draw a ray from the top of the object parallel to the principal axis, which passes through the focal point on the right side of the lens.
4. Draw a ray from the top of the object passing through the optical center of the lens.
5. Draw a ray from the top of the object passing through the focal point on the left side of the lens.
6. These rays will converge at a point on the left side of the lens, which represents the image.

If the object is displaced 15cm towards the optical center of the lens, the ray diagram will be different. In this case, the image will be formed on the same side of the lens as the object, and it will be smaller in size than the object.

Given v= 25 cm , m= -1
-1 indicate that image is real inverted and same size of the object so image is formed by a convex lens.
m=v/u => -1=25/u => u= -25 cm
1/v-1/u=1/f => 1/f=1/25-(-1/25) =(1+1)/25=2/25
f=25/2 =>f = 12.5 cm
when the object displace 15cm towards optic centre
then (25-15) =10 => u= -10 cm so image
1/v = 1/25/2 +( -1/10) =2/25-1/10=4-5/50= -2/50
v= -50/2 = -25 cm
it's indicate that image is virtual erect and larger than object. In second case the image formed same side to object beyond f1.
I think that is right answer but I also a student of class 10 so if it wrong then inform me.