First of all we find out the position of the image. By the position of image we mean the distance of image from the lens.
Here, Object distance, u=-12 cm (it is to the left of lens)
Image distance, v=? (To be calculated)
Focal length, f=`+8` cm (It is a convex lens)
Putting these values in the lens formula:
`1/v-1/u=1/f`
We get: `1/v-1/(-12)=1/8`
or `1/v+1/12=1/8`
or `1/v+1/12=1/8`
`1/v=1/8-1/12`
`1/v=(3-2)/(24)`
`1/v=1/(24)`
So, Image distance, `v=+24` cm
Thus, the image is formed on the right side of the convex lens. Only a real and inverted image is formed on the right side of a convex lens, so the image formed is real and inverted.
Let us calculate the magnification now. We know that for a lens:
Magnification, `m=v/u`
Here, Image distance, v=24 cm
Object distance, u`=-12` cm
So, `m=24/-12`
or `m=-2`
Since the value of magnification is more than 1 (it is 2), so the image is larger than the object or magnified. The minus sign for magnification shows that the image is formed below the principal axis. Hence, the image is real and inverted. Let us calculate the size of the image by using the formula:
`m=h_(2)/h_(1)`
Here, Magnification, m`=-2` (Found above)
Height of object, `h_(1)=+7` cm (Measured upwards)
Height of image, `h_(2)=?` (To be calculated)
Now, putting these values in the above formula, we get:
`-2=h_(2)/(7)`
or `h_(2)=-2 xx 7`
Thus, Height of image, `h_(2)=-14`cm
Thus, the height or size of hte image is 14 cm. The minus sign shows that this height is in the downward direction, that is the image is formed below the axis. Thus, the image is real and inverted.
