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University of Rochester Department of Physics and Astronomy

PHY 123, Spring 2012

2nd Midterm Exam - SOLUTIONS

March 26, 2012

• Time: 1h 10m. • You are only allowed to use a pen (or pencil + eraser) and a

cheat sheet (one side of one letter size—8.5 in × 11 in—sheet).

• Write the relevant work and solutions on the blue examination book. We will not collect scratch papers.

• Begin each problem on a separate sheet of the blue book, clearly label the problem number, and put your name on each examination blue book you use.

___________________ Short questions (a) A marching band composed by high pitched and low pitched instruments is approaching a cross street. Which type of instruments will you hear first? Explain why.

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(b) In attempting to discern distant details, people will sometimes squint (i.e., look with the eyes partly closed). Why does this help? It reduces the spherical aberration and spreading of the image. (c) Suppose white light falls on the two slits of a Young’s double-slit setup, but one slit is covered by a red filter (700 nm) and the other by a blue filter (450 nm). Describe the pattern on the screen. The red light and the blue light coming from the two different slits will have different wavelengths and different frequencies, thus they will not have a constant phase relationship. In order for a double-slit pattern to be produced, the light coming from the slits must be coherent. No distinct double-slit interference pattern will appear. However, each slit will individually produce a “single-slit diffraction” pattern, as discussed in Chapter 35. Problem 1

As shown in the left figure, a small object is located 40 cm from the first of two thin converging lenses of focal lengths 20 cm and 10 cm, respectively. The lenses are separated by l = 30 cm.

(a) Where is located the final image formed by the two-lens system?

(b) If we replace the first lens (f1 = 20 cm) with a thin planoconcave diverging lens made with a material of refractive index n = 1.5, the two-lens system forms a real inverted image 12 cm to the right of the second converging lens, as shown in the right figure. What are the values of the two radii of curvature forming the front surfaces of the planoconcave diverging lens?

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Solution

(a) Applying twice the thin lens equation, we find that the final image formed by the two-lens system is 0.05 m to the right of the second lens. The first lens gives the image at

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that is, 10 cm to the right of the second lens. Let us redefine di1 as equal to 10 cm. This first image becomes the object for the second lens. Remember that—the object distance is positive if the object is on the side of the lens from which the light is coming, otherwise is negative (Giancoli, 33-2).

(b) Applying the thin lens equation with the final image distance and focal length (f2) of the converging lens, we determine the location of the object for the second lens. Subtracting this distance from the separation distance between the lenses gives us the image distance from the first lens. Following the sign convention, inserting this image distance and object distance in the thin lens equation, we calculate the focal length of the

diverging lens. Using the lensmaker’s equation and knowing that one of the two radii of curvature tends to infinite, we calculate the remaining radius of curvature. Problem 2

Calculate the minimum thickness needed for an antireflecting coating (n = 1.38) applied to a glass lens (n = 1.50) in order to eliminate (a) blue (450 nm), or (b) red (720 nm)

1di1

=1f1−1do1

→ di1 = 40 cm

do2 = −di1 →1di2

=1f2−1do2

=1f2−

1(−di1)

→ di2 = 5 cm = 0.05 m

1do2

+1

di2

=1f2

→ do2 =di2 f2

di2 − f2

=12.0cm( ) 10.0cm( )12.0cm −10.0cm

= 60cm

di1 = l − do2 = 30.0cm − 60.0cm = −30.0cm

1do1

+1

di1

=1f1

→ f1 =di1do1

di1 + do1

=−30.0cm( ) 40.0cm( )−30.0cm + 40.0cm

= −1.20 m

1f1

= (n −1)(1R

) → R= − 0.60m

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reflections for light at normal incidence. (Neglect the dependence of n on the wavelength.) Solution

 

 

With respect to the incident wave, the wave that reflects from the top surface of the coating has a phase change of 1 .φ π= With respect to the incident wave, the wave that reflects from the glass n =1.50( ) at the bottom surface of the coating has a phase change

due to both the additional path length and reflection, so φ2 =2πλfilm

⎛

⎝⎜⎞

⎠⎟2t + π . For destructive

interference, the net phase change must be an odd-integer multiple of .π

( )

( ) ( )

net 2 1film

1 1film4 4

film

2 2 2 1

2 1 2 1 , 0, 1, 2, ...

t m

t m m mn

φ φ φ π π π πλ

λλ

⎡ ⎤⎛ ⎞= − = + − = + →⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

= + = + =

The minimum thickness has m = 0, and so 1min 4

film

.tnλ=

(a) For the blue light: ( )( )

1min 4

450nm81.52nm 82nm .

1.38t = = ≈

(b) For the red light: ( )( )

1min 4

700nm126.8nm 130nm .

1.38t = = ≈

Problem 3

A 5,000 lines/cm grating is used to analyze a spectrum that includes an IR line at 800.00 nm and a violet line at 400.00 nm. (a) What are the angular separations between these two spectral lines? (b) Can you resolve a third line added at 800.05 nm? (c) Can you resolve a third line added at 400.05 nm?

1φ π=

( )2 film2 2 tφ λ π π= +

 

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Solution (a) The grating spacing is

𝑑 = !!"""

 𝑐𝑚 = 2×10!!  𝑚. Diffraction angles at different orders are given by

- for the line at 800 nm, two orders are possible

𝑠𝑖𝑛  𝜃! =𝜆𝑑          ⇒          𝑠𝑖𝑛  𝜃! =

0.8×10!!

2×10!! = 0.4

𝑠𝑖𝑛  𝜃! = 2𝜆𝑑          ⇒          𝑠𝑖𝑛  𝜃! = 2

0.8×10!!

2×10!! = 0.8

- for the line at 400 nm, four orders are effectively possible

𝑠𝑖𝑛  𝜃! =𝜆𝑑          ⇒          𝑠𝑖𝑛  𝜃! =

0.4×10!!

2×10!! = 0.2

𝑠𝑖𝑛  𝜃! = 2 !!          ⇒          𝑠𝑖𝑛  𝜃! = 2 !.!×!"

!!

!×!"!!= 0.4 .

𝑠𝑖𝑛  𝜃! = 5𝜆𝑑          ⇒          𝑠𝑖𝑛  𝜃! = 5

0.4×10!!

2×10!! = 1 Thus

 Δ𝜃! = 𝑠𝑖𝑛!!0.4− 𝑠𝑖𝑛!!0.2                          Δ𝜃! = 𝑠𝑖𝑛!!0.8− 𝑠𝑖𝑛!!0.4 (b) The IR lines cannot be resolved. Using the highest available diffraction order (m = 2):

Δ𝜆 = !!= !

!"                        Δ𝜆 = !""

!"""  (!)= 0.08  𝑛𝑚.

(c) The violet lines can be resolved. Using already the second order (m = 2):

Δ𝜆 = !!= !

!"                        Δ𝜆 = !""

!"""  (!)= 0.04  𝑛𝑚.