Let's solve part (a) of the question using the given values.

The formula that relates the speed of light ($c$), wavelength ($\lambda$), and frequency ($f$) of an electromagnetic wave is:

$c = f \cdot \lambda$

Where:

• $c = 3 \times 10^8 \, \text{m/s}$ (speed of light),
• $\lambda = 6.00 \times 10^{-5} \, \text{m}$ (given wavelength).

Rearranging the formula to find the frequency:

$f = \frac{c}{\lambda}$

Substitute the values:

$f = \frac{3 \times 10^8}{6.00 \times 10^{-5}} = 5 \times 10^{12} \, \text{Hz}$

Now, from the table, we see that this frequency falls within the range of infrared radiation, which is from $10^{11} \, \text{Hz}$ to $10^{14} \, \text{Hz}$.

Thus, the type of electromagnetic wave is infrared.

Part (b):

An application of infrared radiation is in thermal imaging cameras or night-vision devices, as infrared radiation is emitted by warm objects.

Would you like further details or have any questions?

Related Questions:

1. How is the frequency of a wave related to its energy?
2. What are other applications of infrared radiation besides thermal imaging?
3. What is the relationship between frequency and wavelength for electromagnetic waves?
4. How does the speed of light in different mediums affect the wavelength and frequency?
5. How do microwaves differ from infrared waves in terms of frequency and application?

Tip:

Always use the formula $c = f \cdot \lambda$ when relating wavelength and frequency for electromagnetic waves in a vacuum.