A1. 1
Correct formula for $\tau$ (up to numerical coefficient)
$$ \tau = \frac{3\hbar }{k a_0^2 e^2} \left( \frac{\lambda_0}{2\pi} \right)^3 $$ 
0.50 

A1. 2 Correct numerical coefficient  0.25 

A1. 3 Numerical answer in the interval $\tau \in (0.5 \cdot 10^{8},\; 4.0 \cdot 10^{8})\,s$ (if the formula is correct up to numerical coefficient).  0.25 

A2. 1 Answer $W_s = N \frac{\hbar \omega}{\tau}$  0.25 

A3. 5 Answer $\Delta t_s = \tau$  0.25 

A4. 1 Answer $W_i = N^2 \frac{\hbar \omega}{\tau}$  0.50 

A5. 1 Answer $\Delta t_i = \tau/N$  0.25 

B1. 1 Answer $\Delta v = v_1 v_2 \approx \frac{c n_2}{n_0^2}\left( E_{m2}^2 E_{m1}^2\right)$  0.50 

B2. 2 Correct formula for parabolic envelope of $E_m^2(t)$  0.10 

B2. 3 It is shown that frequency depends on the coordinate in the wavepocket or the number of the maximum linearly  0.80 

B2. 4 Analytical answer $s = \frac{K \lambda_0}{8 n_2 E_m^2}$  0.80 

B2. 5 Numerical value $s= 8 m$  0.30 

B3. 1 Correct sign $\beta_2<0$  0.50 

B4. 1 Formula for the group velocity $V_g = \frac{1}{\beta_1 + \beta _2(\omega  \omega_0)}$  0.25 

B4. 2 Formula for the distance $l = \frac{(\Delta t_0)^2 }{2 \pi K \beta_2}$  0.50 

B4. 3 Numerical answer $l= 4 m$  0.25 

B5. 2 Formula for diffraction divergence $\theta_d \approx \lambda/ a$  0.20 

B5. 3 Formula for total internal reflection angle $\sin \alpha_c = n_0/(n_0 + n_2 E_m^2)$  0.40 

B5. 4 Formula for the power $W_c = \frac{\pi \varepsilon_0 c^3}{n_2 \omega_0^2}$  0.50 

B5. 5 Numerical answer $W_c \approx 26 MW$  0.40 

C1. 1 Method suggested and correct physical phenomenon is specified  1.00 

C2. 1 Formula for the velocity of the star $v_1 = \frac{2\pi Rm}{TM}$  0.20 

C2. 2 Projection of the velocity on the line of sight $v_1 \cos \theta$  0.10 

C2. 3 Connection between $\Delta \omega/\omega$ and $v/c$  0.10 

C2. 4 Analytical answer $\Delta \omega/\omega = \frac{2\pi R }{T c} \frac{m}{M}\cos \theta$  0.60 

C3. 1 Numerical value of $n$: $n \approx 10$.  0.15 

C3. 2 Correct statement that precision is enough to detect the exoplanet which does not contradict to the previous results  0.10 
