The distance between an object and a screen is 100 cm. A lens can produce real image of the object on the screen for two different positions between the screen and the object. The distance between these two positions is 40 cm. If the power of the lens is close to $$\left(\frac{N}{100}\right)$$D where N is an integer, the value of N is __________

Let the fixed distance between the object and the screen be denoted by $$L$$. We are told that

$$L = 100 \text{ cm}.$$

Suppose the lens is at a distance $$u$$ from the object. Because the screen is on the other side of the lens, the distance of the image from the lens is then

$$v = L - u.$$

The thin-lens formula is first stated:

$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}.$$

Substituting $$v = L - u$$ gives

$$\frac{1}{f} = \frac{1}{u} + \frac{1}{L - u} = \frac{L}{u(L - u)}.$$

Cross-multiplying, we obtain

$$u(L - u) = fL,$$

which simplifies to a quadratic in $$u$$:

$$u^{2} - Lu + fL = 0.$$

This quadratic has two real roots $$u_{1}$$ and $$u_{2}$$ because the lens forms a sharp (real) image on the screen in two different positions. For a quadratic $$x^{2} - Sx + P = 0,$$ the sum of roots is $$S$$ and the product of roots is $$P.$$ Hence

$$u_{1} + u_{2} = L,$$ $$u_{1}u_{2} = fL.$$

The distance between the two lens positions is the absolute difference of the two roots:

$$d = |u_{1} - u_{2}|.$$

Using the identity $$|u_{1} - u_{2}| = \sqrt{(u_{1} + u_{2})^{2} - 4u_{1}u_{2}},$$ we get

$$d = \sqrt{L^{2} - 4fL}.$$

We are given that this separation is

$$d = 40 \text{ cm}.$$

Substituting $$L = 100 \text{ cm}$$ and $$d = 40 \text{ cm}$$ into the expression for $$d$$, we have

$$40 = \sqrt{100^{2} - 4f(100)}.$$

Squaring both sides:

$$1600 = 10000 - 400f.$$

Rearranging for $$f$$:

$$400f = 10000 - 1600,$$ $$400f = 8400,$$ $$f = \frac{8400}{400} = 21 \text{ cm}.$$

To find the power $$P$$ of the lens in dioptres, we recall the definition:

$$P = \frac{1}{f\text{(in metres)}}.$$

Since $$f = 21 \text{ cm} = 0.21 \text{ m},$$ we have

$$P = \frac{1}{0.21} \text{ D} \approx 4.7619 \text{ D}.$$

The problem states that the power is close to $$\left(\dfrac{N}{100}\right)\text{ D},$$ so

$$\frac{N}{100} \approx 4.7619 \quad\Longrightarrow\quad N \approx 4.7619 \times 100 \approx 476.19.$$

Because $$N$$ must be an integer, we take

$$N = 476.$$

So, the answer is $$476$$.