1. Dynamic resistance of a diode

For a diode, the dynamic (AC) resistance at a point on the forward-bias I-V curve is

$r_d = \frac{dV}{dI}$

On a graph of $I$ vs $V$, this is the reciprocal of the slope:

$r_d = \frac{1}{\frac{dI}{dV}}$

So, the steeper the $I$-$V$ curve, the smaller the dynamic resistance.

2. Use the given graph qualitatively

From the forward-bias characteristic of a p-n junction diode, current increases very rapidly with voltage.

At $V = 2\,\text{V}$, the curve is much less steep than at $V = 4\,\text{V}$. Thus,

• slope at $2\,\text{V}$ is smaller
• slope at $4\,\text{V}$ is larger
• therefore dynamic resistance at $2\,\text{V}$ is larger than at $4\,\text{V}$

Hence,

$r_d(2\,\text{V}) : r_d(4\,\text{V}) > 1$

3. Estimate from diode equation

For a forward biased diode,

$I \propto e^{\frac{V}{\eta V_T}}$

and dynamic resistance is

$r_d = \frac{dV}{dI} = \frac{\eta V_T}{I}$

So,

$r_d \propto \frac{1}{I}$

Therefore,

$\frac{r_d(2)}{r_d(4)} = \frac{I(4)}{I(2)}$

From the plotted curve, current at $4\,\text{V}$ is about 5 times the current at $2\,\text{V}$, so

$r_d(2) : r_d(4) = 5 : 1$

4. Check options

• A: $1:2$ ❌
• B: $5:1$ ✅
• C: $1:40$ ❌
• D: $20:1$ ❌

So the correct answer is:

$\boxed{5:1}$

Hence, Option B is correct.