If the length of the latus rectum of an ellipse is 4 units and the distance between a focus and its nearest vertex on the major axis is $$\frac{3}{2}$$ units, then its eccentricity is:

For any ellipse whose major axis lies along the x-axis, let the semi-major axis be denoted by $$a$$, the semi-minor axis by $$b$$ and the eccentricity by $$e$$. We recall two standard facts:

1. The length of the latus rectum is given by the formula $$\text{Latus rectum}= \frac{2b^{2}}{a}.$$

2. The relation between $$a,\;b$$ and $$e$$ is $$b^{2}=a^{2}\left(1-e^{2}\right).$$

Using the second relation inside the first, we obtain a more convenient single formula: $$\text{Latus rectum}= \frac{2\bigl(a^{2}(1-e^{2})\bigr)}{a}=2a\left(1-e^{2}\right).$$

We are told that the length of the latus rectum equals $$4$$ units, so

$$2a\left(1-e^{2}\right)=4.$$

Dividing both sides by $$2$$ gives

$$a\left(1-e^{2}\right)=2.\qquad(1)$$

Next, consider the distance between a focus and its nearest vertex on the major axis. A focus is located at a distance $$ae$$ from the centre, while the nearer vertex is at a distance $$a$$ from the centre in the same direction. Therefore the required distance equals

$$a-ae=a(1-e).$$

The problem states that this distance is $$\dfrac{3}{2}$$ units, so we have

$$a(1-e)=\frac{3}{2}.\qquad(2)$$

Now we possess two equations in $$a$$ and $$e$$, namely (1) and (2). From equation (2) we can express $$a$$ in terms of $$e$$:

$$a=\frac{\dfrac{3}{2}}{1-e}=\frac{3}{2(1-e)}.\qquad(3)$$

Substituting this value of $$a$$ from (3) into equation (1), we get

$$\left(\frac{3}{2(1-e)}\right)\bigl(1-e^{2}\bigr)=2.$$

Since $$1-e^{2}=(1-e)(1+e)$$, the factor $$1-e$$ cancels:

$$\frac{3}{2}\,(1+e)=2.$$

Multiplying both sides by $$2$$ gives

$$3\,(1+e)=4.$$

Expanding the left side,

$$3+3e=4.$$

Subtracting $$3$$ from both sides yields

$$3e=1.$$

Finally, dividing by $$3$$, we obtain the eccentricity:

$$e=\frac{1}{3}.$$

Hence, the correct answer is Option D.