Suppose I is the real image of an object O. Let d be the distance between them. If the image distance is x, the object distance will be (d – x).
Thus, u = – (d – x) and v = + x
Sustituting in the lens formula we have
For a real image, the value of x must be real, i.e., the roots of the above equation must be real. This is possible if
d2 ≥ 4fd
or, d ≥ 4f
Hence, 4f is the minimum distance between the object and its real image formed by a convex lens.
Distance of object from the mirror, u = –30 cm;
Distance of image, v = ?
Focal length, f = + 20 cm
We know, using the mirror formula,
The image is virtual and erect as, it is formed at a distance of 12 cm behind the mirror.
Now, size if the object, O = 0.5 cm
sixe of the image = I
Hence, the height of the image = + 0.2 cm.
The positive sign indicates that the image is erect is smaller in size
Given, a convex lens.
Refractive index of lens, = 1.5
Focal length of lens in air, fa = 18
Refractive index of water,
For the lens in air,
When the lens is immersed in water
Hence, focal length changes from 18 to 32.
The image is formed at a distance of 7.5 cm in front of mirror.
I is the image size and,
O is the object size.
The negative sign indicates that the image is inverted and is diminished in size.