AN OBJECT OF HEIGHT 5 CM IS PLACED PERPENDICULAR TO THE PRINCIPAL AXIS OF A CONCAVE LENS OF FOCAL LENGTH 10 CM .IF THE DISTANCE OF THE OPTICAL CENTRE OF THE LENS ES 20 CM .DETERMINE THE POSITION ,NATURE AND SIZE OF THE IMAGE FORMED USING THE LES FORMULA.

1/f = 1/p + 1/q

f = 10, p = 5. They value and sign of q will clue you into its nature.
Ohh. And there's no need to shout :)

To determine the position, nature, and size of the image formed using the lens formula, we can follow these steps:

Step 1: Determine the focal length of the concave lens.
Given that the focal length (f) of the lens is 10 cm.

Step 2: Determine the object distance (u) from the optical center of the lens.
Given that the distance of the optical center (O) of the lens is 20 cm.

Since the object is placed perpendicular to the principal axis, the object distance (u) is negative. Therefore, u = -20 cm.

Step 3: Determine the image distance (v) from the lens.
Using the lens formula: 1/f = 1/v - 1/u

By substituting the known values, we get: 1/10 = 1/v - 1/-20

Simplifying this equation, we get: 1/10 = (1 + 2)/v

Now, we can solve for v: 10v = 30

Thus, v = 3 cm.

Step 4: Determine the magnification (m).
Using the magnification formula: m = -v/u

By substituting the known values, we get: m = -3/(-20)

Simplifying this equation, we get: m = 3/20.

Step 5: Determine the nature of the image.
Since the magnification (m) is positive, the image formed is virtual.

Step 6: Determine the size of the image.
The size of the image is given by the magnification (m). In this case, since m = 3/20, the image is smaller than the object.

Therefore, the position of the image is 3 cm in front of the lens (virtual), the nature of the image is virtual (magnification is positive), and the size of the image is smaller than the object.