A ray of light with a frequency of 5.0 imes 1014 ext Hz in air strikes a glass prism (n_g = 1.6) at a glancing angle of 60^ iny exto . Calculate (a) the frequency of the light in the prism, (b) the

Step 1: Identify the given values and interpret the angle of incidence.
The glancing angle is $60^\circ$. The angle of incidence $i$ is measured from the normal, so:

$$i = 90^\circ - 60^\circ = 30^\circ$$

Given frequency of light in air, $f_{air} = 5.0 \times 10^{14} \text{ Hz}$.
Given speed of light in air, $c = 3.0 \times 10^8 \text{ m/s}$.
Given refractive index of glass, $n_g = 1.6$.
The refractive index of air, $n_{air} = 1$.

a) Frequency of the light in the prism
The frequency of light remains constant when it passes from one medium to another.

$$f_{prism} = f_{air}$$ $$f_{prism} = 5.0 \times 10^{14} \text{ Hz}$$

The frequency of the light in the prism is $\boxed{\text{}5.0 \times 10^{14} \text{ Hz}}$.

b) Wavelength of the light in air
The relationship between speed, frequency, and wavelength is $c = f \lambda$.

$$\lambda_{air} = \frac{c}{f_{air}}$$

Substitute the values:

$$\lambda_{air} = \frac{3.0 \times 10^8 \text{ m/s}}{5.0 \times 10^{14} \text{ Hz}}$$ $$\lambda_{air} = 0.6 \times 10^{-6} \text{ m}$$ $$\lambda_{air} = 6.0 \times 10^{-7} \text{ m}$$

The wavelength of the light in air is $\boxed{\text{}6.0 \times 10^{-7} \text{ m}}$.

c) Wavelength of the light in the prism
The refractive index is also given by the ratio of wavelengths: $n_g = \frac{\lambda_{air}}{\lambda_{prism}}$.

$$\lambda_{prism} = \frac{\lambda_{air}}{n_g}$$

Substitute the values:

$$\lambda_{prism} = \frac{6.0 \times 10^{-7} \text{ m}}{1.6}$$ $$\lambda_{prism} = 3.75 \times 10^{-7} \text{ m}$$

The wavelength of the light in the prism is $\boxed{\text{}3.75 \times 10^{-7} \text{ m}}$.

d) Angle of refraction
Use Snell's Law: $n_{air} \sin i = n_g \sin r$.

$$1 \cdot \sin(30^\circ) = 1.6 \cdot \sin r$$ $$\sin r = \frac{\sin(30^\circ)}{1.6}$$ $$\sin r = \frac{0.5}{1.6}$$ $$\sin r = 0.3125$$ $$r = \arcsin(0.3125)$$ $$r \approx 18.21^\circ$$

The angle of refraction is $\boxed{\text{}18.21^\circ}$.

e) Speed of the light in the block
The refractive index is defined as $n_g = \frac{c}{v_{prism}}$, where $v_{prism}$ is the speed of light in the prism.

$$v_{prism} = \frac{c}{n_g}$$

Substitute the values:

$$v_{prism} = \frac{3.0 \times 10^8 \text{ m/s}}{1.6}$$ $$v_{prism} = 1.875 \times 10^8 \text{ m/s}$$

The speed of the light in the block is $\boxed{\text{}1.875 \times 10^8 \text{ m/s}}$.