We know that the visible spectrum has wavelengths of 390nm to 700nm. The wavelengths of light around the 700nm range are red.

Using the ROYGBIV acronym, we know that red has the longest wavelength and violet the shortest. While this may seen to be a difficult question asking for the wavelengths of light that correspond to certain colors, this is helpful on the MCAT for estimating answers and may be worth the time learning. The expected range of wavelengths for the red portion of the visible spectrum is 650nm to 700 nm. Keep in mind that, because it has the longest wavelength, red light will also have the lowest frequency.

This question asks us about the particle nature of light and how much energy a photon would contain. From our light equations, we know that , where E is the energy of a single photon, h is Plank’s constant, and f is the frequency of the photon.

From the information in the problem, we need to determine the frequency.

We need to determine the velocity of the light in the solution. We can use the definition of index of refraction to determine this value, along with the speed of light and the index of refraction of the solution.

Now we can compute the frequency.

Substituting the frequency we found, along with Plank’s constant, we can find the energy.

An important distinction for the MCAT is the difference between transverse and longitudinal waves. Although both wave types are sinusoidal, transverse waves oscillate perpendicular to the direction of propagation, while longitudinal waves oscillate parallel to the direction of propagation.

The most common transverse and longitudinal waves are light waves and sound waves, respectively. All electromagnetic waves (light waves, microwaves, X-rays, radio waves) are transverse. All sound waves are longitudinal.

The velocity of a wave can be obtained with the formula , where  is the tension in the string and is the mass per unit length of the string. Since the tension is quadrupled, the velocity will be doubled.

Let's assume that a string with tension and a mass per unit length produces a wave with velocity .

If we increase the tension by a factor of four, we will get the below expression.

We can see that , and we know that .

The first filter polarizes the light horizontally, only allowing light to pass if it is oscillating horizontally. This would decrease the intensity by one-half.

The second filter would polarize the light vertically, only allowing light to pass if it oscillates vertically. The first filter, however, has already blocked all non-horizontal waves, including any vertical waves; thus, there are no remaining waves that can pass through the vertical filter. The light is fully blocked by this combination.

As the wave travels into the less dense medium, it speeds up, bending away from the normal line. The index of refraction tells the ratio of the velocity in a vacuum in relation to the velocity the medium; thus, the velocity will be greater in a medium with a lower index of refraction.

Frequency remains the same regardless of medium, however, since the velocity changes, the wavelength must accommodate this change.

If velocity increases and frequency remains constant, wavelength must also increase.

Finally, a phase shift only occurs when a light ray reflects from the surface of a more dense medium.

The period of a wave is equal to the reciprocal of the frequency:

Respectively, frequency is the reciprocal of period. By definition, the product of two reciprocals is one.

Wave velocity is given by the product of frequency and wavelength:

In the question, we are given the period (waves per second). To find the frequency, we will need to take the reciprocal of the period.

Using the values given in the question, we can find the velocity of the waves. The wavelength is twice the distance between adjacent maxima and minima, making our wavelength two meters.

Work function is the minimum amount of energy, in electron-volts, required to remove an electron from a metal surface. To calculate the energy of a single photon of light, one must use the equation:

In this formula,  is energy in Joules,  is Planck's Constant, and  is the speed of light.

To convert from Joules to eV, one must use the given conversion factor.

Use this value in the energy formula to find the wavelength of the light.

The visible spectrum spans from approximately to , with smaller wavelengths corresponding to violet and blue and larger wavelengths corresponding to red. Even without knowing the exact wavelength correlations in the spectrum, we know that our wavelength is very small and will be found in the violet end of the spectrum.

Incandescent light bulbs produce visible light of all wavelengths. The mix of red, orange, yellow, green, blue, indigo, and violet (ROYGBIV) give the light its characteristic white appearance. For the MCAT, it is important to know the relative wavelengths of light for the visible spectrum (390 – 700nm) and where visiable wavelengths fit into the overall spectrum of electromagnetic radiation. From longest wavelength to shortest, the sequence of wavelengths is listed below.

Radio > Microwaves > Infrared > Visible > Ultraviolet > X-Rays > Gamma Rays