User:WalkingRadiance/Physics Optics Wavelength

$$\frac{\lambda_2}{\lambda_1} = \frac{n_1}{n_2}$$

$$c=\lambda f$$

$$\lambda=c/f$$

$$\lambda_2 n_2=\lambda_1 n_1$$

$$c/f_2 n_2 =c/f_1 n_2$$

$$\frac{c}{f_2}n_2=\frac{c}{f_1}n_1$$

A1. Laser light of frequency 1.0 10 15 Hz is fired into a bucket of water. (Inside the water, the laser

has the same frequency as outside.)

A) If the index of refraction of the water is 1.33, find the wavelength of the laser trasmitted *in*

the water. (Warning: this will be different from the wavelength outside the water; think

carefully about what is different inside the water.)

$$f_1=1.0\cdot 10^{15}$$

$$n_1=1.000293$$

$$c\longmapsto \lambda f$$

$$\lambda$$ is the wavelength and $$f$$ is the frequency in hertz or inverse seconds and

$$299,792,458\frac{m}{s}=(2.2542.2547\times10^{}-7m)\cdot$$

$$\frac{\sin\theta_2}{\sin\theta_1} = \frac{v_2}{v_1} = \frac{n_1}{n_2}$$

The answer to A1 is $$(2.2542.2547\times10^{-7}m) $$ and

$$E=hf=\frac{hc}{\lambda}$$

$${}_{5}^{11}\!B+{}_{1}^{1}\!proton={}_{6}^{12}\!totalmass=3\times {}_{2}^{4}\!He$$

$${}_{proton \#}^{total \# of nucleons}\!nucleons$$

The total number of protons and neutrons in Boron-11 is 11. The total number of protons in boron is 5 because boron's atomic number $$Z$$ is 5. The total number of protons in boron-11 and a proton is 12 (11+1).

$$\frac{\lambda_2}{\lambda_1} = \frac{n_1}{n_2}=\frac{f_1}{f_2}=\frac{\sin(\theta_2)}{\sin(\theta_1)} =\frac{v_2}{v_1}$$ $$\sin(\theta_1)n_1=\sin(\theta_2)n_2$$

The

$$n1=n2=1.000293$$

$$\sin(\theta_1)=\sin(\theta_2)$$

$$\theta_1=\theta_2$$

$$\theta_1+\theta_2+90^\circ=180^\circ$$

The two equations are $$\sin(\theta_1)n_1=\sin(\theta_2)n_2$$, which is simplified to $$\theta_1=\theta_2$$ and $$(\theta_1+\theta_2)+90^\circ=180^\circ$$

$$(\theta_1+\theta_2)+90^\circ=180^\circ$$is subtracted from $$\theta_1+\theta_2=0$$. we now have $$2(\theta_1+\theta_2)$$