Sneaky photons

According to Snell's law: $n_1 \sin \alpha=n_2 \sin \beta$ , so $\beta \approx 35.26^\circ$ .

The change of the direction of vector momentum is: $\phi=\alpha-\beta \approx 24.74^\circ$ .

The change of vector momentum of the photon: $\Delta p=\sqrt{p^2_1+p^2_2-2p_1p_2 \cos \phi}$

$=\frac{h}{\lambda} \sqrt{n^2+1-2n \cos \phi} \approx 865.26 (kgm/s)$ .

The angle of refraction is calculated using Snell's Law: $n_1\sin \theta_1=n_2\sin \theta_2$ , so $\sin \theta_2 = \frac{1\times \sin60^\circ}{1.5}=0.57735$ , giving $\theta_2 = \sin^{-1}0.57735=35.26^\circ$

Now we calculate the magnitudes of the initial and final momenta. The initial momentum is $p_1=\frac{h}{\lambda_1}=\frac{6.63\times 10^{-34}}{555\times10^{-9}} =1.1946 \times 10^{-27} \ \mathrm{kgms^{-1}}$

For the final momentum, note that the speed of the photon decreases by $1.5$ times, and since $v=f\lambda$ while the frequency $f$ remains constant, the wavelength $\lambda$ must decrease by $1.5$ times. So we have $p_2=\frac{h}{\lambda_2}=\frac{h}{\lambda_1/1.5} =1.5p_1=1.7918 \times 10^{-27} \ \mathrm{kgms^{-1}}$

Since momentum is a vector quantity, we cannot simply subtract the magnitudes to find the vector change in momentum. We will have to consider two components - parallel to the surface and perpendicular to the surface.

$\Delta p_{//} =p_2\sin\theta_2 - p_1\sin\theta_1=-1.12096 \times 10^{-31} \ \mathrm{kgms^{-1}}$ $\Delta p_{\perp} =p_2\cos\theta_2 - p_1\cos\theta_1=8.6586 \times 10^{-28} \ \mathrm{kgms^{-1}}$

And so the magnitude of the change in momentum is $|\Delta p| = \sqrt{\Delta p_{//}^2 + \Delta p_{\perp}^2}= 8.65 \times 10^{-28} \ \mathrm{kgms^{-1}}$

The answer will be $\fbox{865}$