Brock University
Physics 1P22/1P92
Winter 2014
Dr. D’Agostino
Solutions for Tutorial 4: Wave Optics (Chapter 17)
1. A 5.0-cm-thick layer of oil (n = 1.46) is sandwiched between a 1.0-cm-thick sheet of
glass (n = 1.50) and a 2.0-cm-thick sheet of polystyrene (n = 1.59). How long (in ns)
does it take light incident perpendicular to the glass to pass through this 8.0-cm-thick
sandwich? [4 points]
Solution: Determine the time needed for light to pass through each of the three layers,
and then add the three times to obtain the total time.
In each layer, the speed of light is obtained using v = c/n. Thus,
t1 =
d1 n1
(5.0 × 10−2 )(1.46)
d1
=
=
= 0.243 ns
v1
c
3.0 × 108
t2 =
d2 n2
(1.0 × 10−2 )(1.50)
=
= 0.05 ns
c
3.0 × 108
d3 n3
(2.0 × 10−2 )(1.59)
=
= 0.106 ns
c
3.0 × 108
The total travel time is therefore t1 + t2 + t3 = 0.4 ns.
t3 =
2. A 500 line/mm diffraction grating is illuminated by light of wavelength 510 nm. How
many diffraction orders are seen, and what is the angle of each? [4 points]
Solution: From the given information that there are 500 lines per millimetre, we can
determine that the spacing between the lines is
d=
1 mm
= 2 × 10−6 m = 2000 nm
500
The angle of each diffraction line satisfies
sin θm =
mλ
510 nm
=m
= 0.255m
d
2000 nm
⇒
θm = sin−1 (0.255m)
Using a calculator, it follows that
θ0
θ1
θ2
θ3
θ4
= sin−1 (0.255(0)) = 0
= sin−1 (0.255(1)) = 14.77◦
= sin−1 (0.255(2)) = 30.66◦
= sin−1 (0.255(3)) = 49.9◦
= sin−1 (0.255(4)) = DOES NOT EXIST
There are three orders of bright lines on each side of the central bright line.
3. A sheet of glass is coated with a 500-nm-thick layer of oil (n = 1.42). [4 points]
For which visible wavelengths of light do the reflected waves interfere (a) constructively? (b) destructively?
(c) What is the colour of reflected light?
(d) What is the colour of transmitted light?
Solution: Assume that the glass has an index of refraction n = 1.50 (see table on
Page 547 of textbook). Then as light from the air reflects from the air-oil interface, it
undergoes a phase change. Similarly, when light from inside the oil reflects from the
oil-glass interface, it also undergoes a phase change. Thus, for constructive interference
of the reflected waves, the path difference must be a whole number of wavelengths:
λm
2t = m
noil
where t is the thickness of the oil layer. The wavelengths of light that work satisfy
2tnoil
2(500)(1.42)
1420 nm
λm =
=
=
m
m
m
Using a calculator, the first few wavelengths can be determined:
λ1 = 1420 nm
1420
= 710 nm
λ2 =
2
1420
λ3 =
= 473 nm
3
1420
λ4 =
= 355 nm (etc.)
4
The only wavelength in the visible range is the 473 nm wave, which is blue. (The 710
nm wave is slightly in the infrared region, but it might be barely visible as red light.)
Thus, the colour of reflected light is blue.
On the other hand, for destructive interference of the reflected waves, the path difference should be a “half-integer” multiple of the wavelength:
2(500)(1.42)
1420 nm
2tnoil
λm =
=
1 =
1
m+ 2
m+ 2
m + 21
Using a calculator, the first few wavelengths can be determined:
1420
λ0 =
= 2840 nm
0.5
1420
= 947 nm
λ1 =
1.5
1420
λ2 =
= 568 nm
2.5
1420
λ3 =
= 406 nm
3.5
and so on. Waves with wavelength 406 nm (violet) and 568 nm (green) are visible.
Thus, the colour of transmitted light is violet and green.