Electronic – n equation for the relationship between illuminance and the resistance of a light dependent resistor

The value of \$r_{o}\$ depends on the transistor's characteristics and its biasing:

$$r_{o} = \frac{V_{\text{A}} + V_{\text{CE}}}{I_{\text{C}}} \approx \frac{V_{\text{A}}}{I_{\text{C}}}$$

where \$V_{\text{A}}\$ is the transistor's Early voltage and can vary widely between transistors, but is usually on the order of \$10 - 100\text{ V}\$. Assuming a typical bias of \$I_{\text{C}} \approx 1\text{ mA}\$, \$r_{o}\$ should be on the order of \$10 - 100\text{ k}\Omega\$. If \$R_{\text{L}} = 1\text{ k}\Omega\$ and \$r_{o}\$ is only \$10\text{ k}\Omega\$ then you could have a significant error if you assume \$r_{o} \parallel R_{\text{L}} \approx R_{\text{L}}\$. But if \$V_{\text{A}}\$ is higher or \$I_{\text{C}}\$ is lower then it may be safe to ignore \$r_{o}\$, depending on how accurate you are looking to be.

Also, it's worth noting that for calculating \$R_{\text{in}}\$ the resistance \$R_{\text{E}} = r_{o}||R_{\text{L}}\$ at the emitter is magnified by \$\beta + 1\$ so an error in \$R_{\text{E}}\$ due to ignoring \$r_{o}\$ could be magnified significantly. On the other hand, for calculating \$R_{\text{out}}\$ the emitter resistance \$R_{\text{E}}\$ is divided by \$\beta + 1\$ so the error would not be as significant.

1) The DC "resistance" is simply \$V/I\$, which in your case is \$0.7\cdot2k / (10 - 0.7) = 150\Omega\$. The dynamic resistance is \$\frac{dV}{dI}\$ of the diode, which can be easily found on the diode datasheet. Like this, for example:

_{(source: davidbridgen.com)}

Here \$\frac{dV}{dI}\$ is the inverse of the slope of the line around the \$0.7V\$ point. As you can see the line is almost vertical, so the slope is very big. So the inverse is very small.

2) The voltage across the diode is not \$0.01V\$. It is \$0.7\pm 0.01V\$. But since the dynamic resistance is the derivative, the DC component is going away.