Physics 41 Chapter 38 HW Key

1. Helium–neon laser light (632..8 nm) is sent through a 0.300-mm-wide single slit. What is the width

of the central maximum on a screen 1.00 m from the slit?

73

46.328 10sin 2.11 103.00 10a

λθ−

−−

×= = = ×

× tan sin

1.00 my

θ θ θ= ≈ ≈ (for small θ)

2 4.22 mmy =

2. A beam of monochromatic green light is diffracted by a slit of width 0.550 mm. The diffraction

pattern forms on a wall 2.06 m beyond the slit. The distance between the positions of zero intensity on both sides of the central bright fringe is 4.10 mm. Calculate the wavelength of the light.

P38.2 The positions of the first-order minima are sinyL a

λθ≈ = ± . Thus, the spacing between these two minima is

2y Laλ ∆ =

and the wavelength is

3 34.10 10 m 0.550 10 m 547 nm

2 2 2.06 my a

Lλ

− − ∆ × × = = =

Coherent light of wavelength 501.5 nm is sent through two parallel slits in a large, flat wall. Each slit is 0.700 µ m wide. Their centers are 2.80 µ m apart. The light then falls on a semicylindrical screen, with its axis at the midline between the slits. (a) Predict the direction of each interference maximum on the screen as an angle away from the bisector of the line joining the slits. (b) Describe the pattern of light on the screen, specifying the number of bright fringes and the location of each. (c) Find the intensity of light on the screen at the center of each bright fringe, expressed as a fraction of the light intensity Imax at the center of the pattern. Solution: First some review: The total intensity of the double slit interference pattern is a combination of single slit diffraction and the double slit interference. The cosine squared term gives the intensity of the fringes inside the single slit maxima. The sine squared terms is the ‘envelope’ that shapes the fringes.

Without the diffraction envelope all of the fringes would have equal intensity. This is what we were shown in chapter 37, although this pattern doesn’t really exist.

( )/I I

/ =

2

2max

sin sin sin cos sin

πa θ λπd θλ πa θ λ

So for this problem, you use the full expression to find the intensity of the bright fringes. The single slit minima DOMINATE the double slit interference so if a double maxima falls on a single slit minima, the spot will be dark.

To find the location of the fringes: Double-slit interference maxima are at angles given by sind mθ λ= . Single-slit diffraction minima, at sina mθ λ= .

(a) The ‘semicylindrical screen’ is irrelevant. The point is, they include that because it would be difficult to SEE fringes at large angles unless the screen was curved – but it doesn’t matter when you are solving for the angles.

Double-slit interference maxima are at angles given by sind mθ λ= .

For 0m = , 0 0θ = °

For 1m = , ( ) ( )2.80 m sin 1 0.501 5 mµ θ µ= : ( )11 sin 0.179 10.3θ −= = °

Similarly, for 2, 3, 4, 5m = and 6, 2 21.0θ = ° , 3 32.5θ = ° , 4 45.8θ = °

5 63.6θ = ° , and ( )16 sin 1.07θ −= = nonexistent

Thus, counting the central fringe, there are 5 5 1 11 directions for interference maxima+ + = . (b) We check for missing orders by looking for single-slit diffraction minima, at sina mθ λ= because

the single slit diffraction DOMINATES the double slit interference:

For 1m = , ( ) ( )0.700 m sin 1 0.501 5 mµ θ µ= and 1 45.8θ = °

Thus, there is no bright fringe at this angle. There are only nine bright fringes , at

0 , 10.3 , 21.0 , 32.5 , and 63.6θ = ° ± ° ± ° ± ° ± ° .

c)

To find the ratio of intensities, you have to use this:

( )/I I

/ =

2

2max

sin sin sin cos sin

πa θ λπd θλ πa θ λ

The teacher’s solution is using this equation too but they don’t explain HOW they use it. That is why reverse engineering the teacher’s solution manual can be a waste of time. Let me explain their solution. In part (a) we found angles for which sind mθ λ= . If you put those angles into this equation, the cosine squared term will always be equal to 1. We solved for the maxima. Where the cosine squared term is equal to 1. SO that term drops out and we just end up with using:

( ) 2

maxsin sin

sina

I Iaπ θ λ

π θ λ

=

Now what they are doing for the rest of the solution is just solving for the argument and making the units work out because you can’t have use both radian and degrees at the same time in your calculator. Check it out. If I had just one big equation, see what happens. I’m going to substitute in the givens and simplify, leaving only the angle:

( ) ( )26 6 2

6 6max

sin (.7 10 )sin (.5015 10 ) sin 4.38507sin4.38507sin(.7 10 )sin (.5015 10 )

x m x mII x m x m

π θ θθπ θ

− −

− −

= =

Check it out – if you put in an angle, the top argument in the sine is then in radians and you can’t have

both radians and angles at the same time in your calculator. So the easiest way I THINK to do this is to make a substitution:

( ) 2

max

sinII

ββ

=

where sinaβ π θ λ=

Then for each angle, solve for β in degree measure and then switch your calculator to radian and put β

in and solve for the final answer.

At 0θ = ° , lim sin 1θθ

→ andmax

1.00II

→

At 10.3θ = ° , use degrees: β =( )0.700 m sin10.3sin 0.78406

0.501 5 ma π µπ θλ µ

°= = which is RADIANS.

Now switch your calculator to radians:

( ) 2

max

sin 0.784060.8107

0.78406I

I

= =

OR, if you don’t want to switch to radians, then you have to convert β into degrees for the top and then you can

keep your calculator in degrees:

( )0.700 m sin 10.3sin0.785 rad 45.0

0.501 5 ma π µπ θλ µ

°= = = °

2

max

sin 45.0 0.8110.785

II

° = =

And that is what the teacher solution manual did. It works either way but for this intensity formula you can’t use both radians and degrees at the same time!!!! So, moral of the story, maybe it is better not to have the teacher solution manual!!!!

Similarly, at 21.0θ = ° , sin1.57 rad 90.0

aπ θλ

= = °andmax

0.405II

=

At 32.5θ = ° , sin2.36 rad 135

aπ θλ

= = ° andmax

0.090 1II

=

At 63.6θ = ° , sin3.93 rad 225

aπ θλ

= = ° andmax

0.032 4II

=

3. The pupil of a cat’s eye narrows to a vertical slit of width 0.500 mm in daylight. What is the angular resolution for horizontally separated mice? Assume that the average wavelength of the light is 500 nm.

73

45.00 10 msin 1.00 10 rad

5.00 10aλ

θ−

−−

×= = = ×

×

4. The Impressionist painter Georges Seurat created paintings with an enormous number of dots of pure pigment, each of which was approximately 2.00 mm in diameter. The idea was to have colors such as red and green next to each other to form a scintillating canvas. Outside what distance would one be unable to discern individual dots on the canvas? (Assume that λ = 500 nm and that the pupil diameter is 4.00 mm.) Sunday Afternoon on the Island of La Grande Jatte, by Georges Seurat.

By Rayleigh’s criterion, two dots separated center-to-center by 2.00 mm would overlap

when min 1.22dL D

λθ = = .

Thus, ( ) ( )

( )3 3

9

2.00 10 m 4.00 10 m13.1 m

1.22 1.22 500 10 mdDL

λ

− −

−

× ×= = =

×.

5. A circular radar antenna on a Coast Guard ship has a diameter of 2.10 m and radiates at a frequency of 15.0 GHz. Two small boats are located 9.00 km away from the ship. How close together could the boats be and still be detected as two objects?

P38.18 1.22 dD Lλ= 0.020 0 mc

fλ = =

2.10 mD = 9 000 mL =

( ) ( )0.020 0 m 9 000 m

1.22 105 m2.10 m

d = =

6. The hydrogen spectrum includes a red line at 656 n and a blue-violet line at 434 nm. What are the

angular separations between these two spectral lines obtained with a diffraction grating that has 4 500 grooves/cm?

P38.21The grating spacing is 2

61.00 10 m 2.22 10 m4 500

d−

−×= = ×

In the 1st-order spectrum, diffraction angles are given by

sindλθ = :

9

1 6656 10 msin 0.2952.22 10 m

θ−

−×

= =×

so that for red 1 17.17θ = ° and for blue 9

2 6434 10 msin 0.1952.22 10 m

θ−

−×

= =×

so that 2 11.26θ = °

The angular separation is in first-order, 17.17 11.26 5.91θ∆ = °− ° = °

In the second-order spectrum, 1 11 22 2sin sin 13.2d dλ λ

θ − − ∆ = − = °

Again, in the third order,

1 11 23 3sin sin 26.5d dλ λ

θ − − ∆ = − = °

Since the red does not appear in the fourth-order spectrum, the answer is complete.

7. A helium–neon laser (λ = 632.8 nm) is used to calibrate a diffraction grating. If the first-order maximum occurs at 20.5°, what is the spacing between adjacent grooves in the grating?

P38.22 sin 0.350θ = : 3632.8 nm 1.81 10 nmsin 0.350

d λθ

= = = ×

Line spacing 1.81 mµ=

8. A diffraction grating has 4 200 rulings/cm. On a screen 2.00 m from the grating, it is found that for a particular order m, the maxima corresponding to two closely spaced wavelengths of sodium (589.0 nm and 589.6 nm) are separated by 1.59 mm. Determine the value of m.

P38.29 61 2.38 10 m 2 380 nm4 200 cm

d −= = × =

sind mθ λ= or 1sin mdλθ − =

and

1tan tan sin my L Ldλθ − = =

Thus, 1 12 1tan sin tan sinm my Ld dλ λ− − ∆ = −

For 1m = ,

( ) 1 1589.6 5892.00 m tan sin tan sin 0.554 mm2 380 2 380

y − − ∆ = − =

For 2m = ,

( ) ( ) ( )1 12 589.6 2 5892.00 m tan sin tan sin 1.54 mm

2 380 2 380y − −

∆ = − =

For 3m = ,

( ) ( ) ( )1 13 589.6 3 5892.00 m tan sin tan sin 5.04 mm

2 380 2 380y − −

∆ = − =

Thus, the observed order must be 2m = .

9. Unpolarized light passes through two ideal Polaroid sheets.The axis of the first is vertical and that of the second is at 30.0° to the vertical. What fraction of the incident light is transmitted?

P38.34 The average value of the cosine-squared function is one-half, so the first polarizer transmits 12

the light. The second

transmits 2 3cos 30.04

° = . 1 3 32 4 8f i iI I I= × =

10. The critical angle for total internal reflection for sapphire surrounded by air is 34.4°. Calculate the polarizing angle for sapphire.

P38.37 1sin c nθ = or 1 1 1.77

sin sin 34.4cn

θ= = =

°.

Also, tan p nθ = . Thus, ( ) ( )1 1tan tan 1.77 60.5p nθ − −= = = ° .

11. The Very Large Array (VLA) is a set of 27 radio telescope dishes in Caton and Socorro Counties, New Mexico (Fig. P38.54). The antennas can be moved apart on railroad tracks, and their combined signals give the resolving power of a synthetic aperture 36.0 km in diameter. (a) If the detectors are tuned to a frequency of 1.40 GHz, what is the angular resolution of the VLA? (b) Clouds of hydrogen radiate at this frequency. What must be the separation distance of two clouds at the center of the galaxy, 26 000 lightyears away, if they are to be resolved? (c) What If? As the telescope looks up, a circling hawk looks down. Find the angular resolution of the hawk’s eye. Assume that that the hawk is most sensitive to green light having a wavelength of 500 nm and that it has a pupil of diameter 12.0 mm. (d) A mouse is on the ground 30.0 m below. By what distance must the mouse’s whiskers be separated if the hawk can resolve them?

P38.50 (a) vf

λ = : 8

9 13.00 10 m s

0.214 m1.40 10 s

λ −

×= =

×

min 1.22Dλ

θ = : min 40.214 m1.22 7.26 rad

3.60 10 mθ µ = = ×

min180 60 60 s7.26 rad 1.50 arc secondsθ µ

π× × = =

(b) mindL

θ = : ( ) ( )6min 7.26 10 rad 26 000 ly 0.189 lyd Lθ −= = × =

(c) min 1.22Dλ

θ = ( )9

min 3500 10 m1.22 50.8 rad 10.5 seconds of arc12.0 10 m

θ µ−

−

×= = ×

(d) ( ) ( )6 3min 50.8 10 rad 30.0 m 1.52 10 m 1.52 mmd Lθ − −= = × = × =

12. An American standard television picture is composed of about 485 horizontal lines of varying light intensity. Assume that your ability to resolve the lines is limited only by the Rayleigh criterion and that the pupils of your eyes are 5.00 mm in diameter. Calculate the ratio of minimum viewing distance to the vertical dimension of the picture such that you will not be able to resolve the lines. Assume that the average wavelength of the light coming from the screen is 550 nm.

P38.57 The limiting resolution between lines ( )( )

94

min 3

550 10 m1.22 1.22 1.34 10 rad

5.00 10 mDλ

θ−

−−

×= = = ×

×.

Assuming a picture screen with vertical dimension , the minimum viewing distance for no visible lines is found from

min485L

θ =

. The desired ratio is then

( )4

min

1 1 15.4485 485 1.34 10 rad

Lθ −

= = =×

.

13. If the interplanar spacing of NaCl is 2.814 × 10–10 m, what is the predicted angle at which X-rays of

wavelength 0.140 nm will be diffracted in a first-order maximum? Answer: 14.4°

14. Suppose a 5-meter diameter telescope were constructed on the dark side of the moon. The viewing there

(except for brief periods of sunlight) would be excellent. As an example, what would be the separation between two objects that could just be resolved on the planet Mars in 500 nm light? [The distance to Mars at closest approach is 50 million miles].

Answer: 9.8 km

15. Suppose a 5-meter diameter telescope were constructed on the dark side of the moon. The viewing there (except for brief periods of sunlight) would be excellent. As an example, what would be the separation between two objects that could just be resolved on the planet Mars in 500 nm light? [The distance to Mars at closest approach is 50 million miles]. 16. An unpolarized beam of light is incident on a stack ofideal polarizing filters. The axis of the first filter is perpendicular to the axis of the last filter in the stack. Find the fraction by which the transmitted beam’s intensity is reduced in the following three cases. (a) Three filters are in the stack, each with its transmission axis at 45.0° relative to the preceding filter. (b) Four filters are in the stack, each with its transmission axis at 30.0° relative to the preceding filter. (c) Seven filters are in the stack, each with its axis at 15.0° relative to the preceding filter.(d) Comment on comparing the answers to parts (a), (b), and (c). 17. Monochromatic light is beamed into a Michelson interferometer. The movable mirror is displaced 0.382 mm, causing the interferometer pattern to reproduce itself 1 700 times. Determine the wavelength of the light. What color is it?