Number the lamps from left to right:

\[ \begin{align*} L_1,\;L_2,\;L_3,\;L_4,\;L_5. \end{align*} \]

After the switch is closed, the outer lamps \(L_1\) and \(L_5\) light up brightly. The inner lamps \(L_2\), \(L_3\), and \(L_4\) remain dark or almost dark.

The reason is that some diodes become forward-biased and provide low-voltage bypass paths around parts of the lamp chain.

In particular:

• one diode bypasses the middle part of the circuit,
• another diode bypasses one of the inner lamps,
• therefore the voltage across these bypassed lamps is too small to make them light brightly.

With the given circuit values, the voltage across the leftmost and rightmost lamps is clearly larger than \(5~{\rm V}\), while the voltage across the bypassed inner lamps is far below \(5~{\rm V}\).

So the result is:

\[ \begin{align*} \boxed{ L_1 \text{ and } L_5 \text{ light up brightly.} } \end{align*} \]

\[ \begin{align*} \boxed{ L_2,\;L_3,\;L_4 \text{ remain dark.} } \end{align*} \]

Both diodes are forward-biased.

The voltage after \(D_1\) is

\[ \begin{align*} U_{R1} = U_0-U_{\rm F} = 4.0~{\rm V}-0.6~{\rm V} = 3.4~{\rm V}. \end{align*} \]

Therefore the current through \(R_1\) is

\[ \begin{align*} I_{R1} = \frac{U_{R1}}{R_1} = \frac{3.4~{\rm V}}{200~\Omega} = 17~{\rm mA}. \end{align*} \]

The voltage after \(D_2\) is

\[ \begin{align*} U_{R2} = U_0-2U_{\rm F} = 4.0~{\rm V}-2\cdot 0.6~{\rm V} = 2.8~{\rm V}. \end{align*} \]

Therefore the current through \(R_2\) is

\[ \begin{align*} I_{R2} = \frac{U_{R2}}{R_2} = \frac{2.8~{\rm V}}{100~\Omega} = 28~{\rm mA}. \end{align*} \]

The current through \(D_1\) supplies both branches:

\[ \begin{align*} I_{D1} = I_{R1}+I_{R2} = 17~{\rm mA}+28~{\rm mA} = 45~{\rm mA}. \end{align*} \]

Thus:

\[ \begin{align*} \boxed{ I_{R1}=17~{\rm mA} } \end{align*} \]

\[ \begin{align*} \boxed{ I_{R2}=28~{\rm mA} } \end{align*} \]

\[ \begin{align*} \boxed{ I_{D1}=45~{\rm mA} } \end{align*} \]

With the switch open, only \(D_1\) is connected to the resistor path.

The node voltage is clamped to approximately

\[ \begin{align*} U_{\rm node} \approx U_{\rm F} = 0.7~{\rm V}. \end{align*} \]

Thus the current through \(R_1\) is

\[ \begin{align*} I_{R1} = \frac{U_0-U_{\rm F}}{R_1} = \frac{5.0~{\rm V}-0.7~{\rm V}}{1.0~{\rm k}\Omega} = 4.3~{\rm mA}. \end{align*} \]

Since only \(D_1\) conducts,

\[ \begin{align*} I_{D1}=4.3~{\rm mA}, \qquad I_{D2}=0. \end{align*} \]

So for open switch:

\[ \begin{align*} \boxed{ S \text{ open: } I_{R1}=4.3~{\rm mA},\quad I_{D1}=4.3~{\rm mA},\quad I_{D2}=0 } \end{align*} \]

With the switch closed, \(D_1\) and \(D_2\) are connected in parallel. The resistor current remains

\[ \begin{align*} I_{R1} = \frac{5.0~{\rm V}-0.7~{\rm V}}{1.0~{\rm k}\Omega} = 4.3~{\rm mA}. \end{align*} \]

However, with the ideal constant-voltage diode model, the individual currents through two parallel diodes are not uniquely determined. The model says only that the sum is

\[ \begin{align*} I_{D1}+I_{D2}=4.3~{\rm mA}. \end{align*} \]

If both diodes are assumed to be identical real diodes, the current would approximately split equally:

\[ \begin{align*} I_{D1} \approx I_{D2} \approx \frac{4.3~{\rm mA}}{2} = 2.15~{\rm mA}. \end{align*} \]

So for closed switch:

\[ \begin{align*} \boxed{ S \text{ closed: } I_{R1}=4.3~{\rm mA}, \quad I_{D1}+I_{D2}=4.3~{\rm mA} } \end{align*} \]

and for approximately identical real diodes:

\[ \begin{align*} \boxed{ I_{D1}\approx I_{D2}\approx 2.15~{\rm mA}. } \end{align*} \]