OPTI 201R Homework 1
Due: September 6, 2016
Homework is due in class. Do all problems and show your work. Credit is not given for answers only.
You are welcome to work together, but be sure your homework is your work.
1. The human eye is sensitive to wavelengths from approximately 400 to 700 nm. What frequency range
does this correspond to?
Frequency (ν) is given by the equation: 𝑣𝑣 =
𝑐𝑐
𝜆𝜆
where λ is in meters and c is in meters/sec.
This results in a range of frequencies from 428.57 THz to 750 THz or 4.2857 x 1014 Hz to
7.5 x 1014 Hz.
2. Your cell phone transmits and receives radio waves at frequencies between 1850 and 1990 MHz.
What wavelength range does this correspond to?
We can rearrange the equation we used in Problem 1 to solve for the wavelength resulting in the
𝑐𝑐
following equation: 𝜆𝜆 = . Remember that the resulting calculation will give a result having
𝑣𝑣
units of meters if c is in meters/sec and ν is in Hz. (Note that 1 MHz = 1 x 106 𝐻𝐻𝐻𝐻). Performing
this calculation results in a wavelength range from 0.1508 meters to 0.1622 meters or 15.08 cm
to 16.22 cm.
3. The website http://refractiveindex.info/ is an online database of refractive indices for various
materials. SF5 is a common glass used in optical elements. Use the database to find the refractive index
of SF5 for a wavelength of λ = 0.5876 µm. What is the speed of light within SF5?
The refractive index is n = 1.6727. The speed of light within SF5 is
𝑐𝑐
𝑛𝑛
=
3×108 𝑚𝑚/𝑠𝑠
1.6727
= 1.79 × 108 𝑚𝑚/𝑠𝑠.
4. Points A and B are separated by three glass plates. For light traveling from A to B, answer the
questions below. The indices of refraction and thicknesses are given in the figure below.
a) What is the physical distance between A and B?
The physical distance between A and B is the sum of all the distances (12 mm +10 mm +15 mm
+5 mm +12 mm) which is 54 mm.
b) What is the optical path length (OPL) between A and B?
Optical path length (OPL) is given by the equation: 𝑂𝑂𝑂𝑂𝑂𝑂 = 𝑛𝑛 ∗ 𝑑𝑑. As in part a) we must sum all
of the distances but in this case we must take into consideration the refractive index of each
length. The total OPL is then equal to 1*12mm + 1.6*10 mm +1.4*15 mm +1.5*5 mm +1*12 mm
which results in a total OPL of 68.5 mm.
c) Where does the light travel slowest?
The speed of the light can be calculated for each region similar to Problem 3. Performing this
calculation for the speed of the light in each medium the following result can be obtained.
n
V*10^8
1
3
1.6
1.875
1.4
2.1428571
1.5
2
1
3
The result is that the light travels slowest in the region of highest index, or in this case when
n=1.6.
d) Where does the light travel fastest?
Similar to part c) of this problem, we can again calculate the speed of light for each medium and
again obtain the same table that was obtained in part c). The light travels fastest in the region
of the lowest index, or in this case when n=1.
e) How does the OPL change if I shuffle the order of the glass plates?
The OPL does not change as long as the thickness of each plate and the total physical distance
does not change. This can be seen in the calculation of OPL in part b), since OPL is a summation
of distances the order does not matter as long as the glass plates do not change index or
thickness.