What is the distance between an object and its real image formed by a thin converging lens with focal length f = 21 cm, if the object distance is 100 cm? (b) What is the minimum object-image distance for a real image formed by that lens?

1/do+1/di=1/f

1/di= 1/0.21 – 1/1 = 3.76
di=0.26 m
D=di+do =0.26+1 = 1.26 m

(b)
1/do+1/di=1/f
di =f•do/(do-f)
D=do+di =
=do+ f•do/(do-f) =do²/(do-f)
For finding D(min) let differentiate D(do)
D´={do²/(do-f)}´={2•do(do-f) -do²(1-f)}/(do-f) ²
D´=0 =>
2•do²- 2•do•f-do²+do²•f=0
do(do-do•f-2•f) =0
d0(1-f) = 2f
d0 =2f/(1-f) = 2•0.21/(1-0.79) = 0.532 m
D(min) = do+di =do + f•do/(do-f)=
=0.532 +{0.21•0.53/(0.53-0.241) =
=0.532+0.348 = 0.88 m

To find the distance between an object and its real image formed by a thin converging lens, we can use the lens formula:

1/f = 1/v - 1/u

Where:
- f is the focal length of the lens,
- v is the image distance from the lens (positive if the image is real and on the opposite side of the lens),
- u is the object distance from the lens (positive if the object is on the same side as the incident light).

a) Distance between object and real image:
Given:
- f = 21 cm (focal length)
- u = 100 cm (object distance)

Substituting the values into the lens formula, we get:

1/21 = 1/v - 1/100

To solve for v, we can rearrange the equation:

1/v = 1/21 + 1/100
1/v = (5 + 1) / 100
1/v = 6/100
v = 100/6
v ≈ 16.67 cm

Therefore, the distance between the object and its real image is approximately 16.67 cm.

b) Minimum object-image distance for a real image:
The minimum object-image distance occurs when the object is placed at the focal point of the lens. In this case, the object distance (u) will be equal to the focal length (f).

Given:
- f = 21 cm (focal length)

Therefore, the minimum object-image distance for a real image formed by this lens is 21 cm.