10. (a) The problem indicates that we should ignore the decrease in sound amplitude
which means that all waves passing through point P have equal amplitude. Their
superposition at P if d = /4 results in a net effect of zero there since there are four
sources (so the first and third are /2 apart and thus interfere destructively; similarly for
the second and fourth sources).
(b) Their superposition at P if d = /2 also results in a net effect of zero there since there
are an even number of sources (so the first and second being /2 apart will interfere
destructively; similarly for the waves from the third and fourth sources).
(c) If d =  then the waves from the first and second sources will arrive at P in phase;
similar observations apply to the second and third, and to the third and fourth sources.
Thus, four waves interfere constructively there with net amplitude equal to 4sm.
11- 20 degrees clockwise from the y axis for both
12- 30. (a) The upward force supplied by radiation pressure in this case (Eq. 33-32) must
be equal to the magnitude of the pull of gravity (mg). For a sphere, the “projected” area
(which is a factor in Eq. 33-32) is that of a circle A = r2 (not the entire surface area of
the sphere) and the volume (needed because the mass is given by the density multiplied
4
by the volume: m = V) is V = 3 r3. Finally, the intensity is related to the power P of
the light source and another area factor 4R2, given by Eq. 33-27. In this way, with  =
19000 in SI units, equating the forces leads to
4
P = 4R2 c  3 r3g /r2 = 4.68  1011 W.
(b) Any chance disturbance could move the sphere from being directly above the source,
and then the two force vectors would no longer be along the same axis.
13- d’=(f2/f1)d
14- 125. (a) For the image formed by the first lens
F1 1 I F 1  1 IJ  20 cm.
G J G
Hf p K H10 cm 20 cmK
1
i1
1
1
1
For the subsequent image formed by the second lens p2 = 30 cm – 20 cm = 10 cm, so
F1 1 I F 1  1 IJ  50 cm.
G J G
Hf p K H12.5 cm 10 cmK
1
i2
2
2
1
Thus, the final image is 50 cm to the left of the second lens, which means that it coincides
with the object.
(b) The magnification is
m
F
i IF
i I F
20 cmIF
50 cmI
 G JG
G
J
G
J
Hp KHp K H20cmKH10cm J
K 5.0,
1
1
2
2
which means that the final image is five times larger than the original object.
(c) The image is virtual since i2 < 0.
(d) The image is inverted since m < 0.
15- I did this one in class for sound waves. Solution is the same.
16- No phase difference due to reflection as there are 2 reflections.
N1=nL/λ; N2=2nL/λ; Ф=2π(N2-N1) =2 π n L/ λ.
For destructive interference: 2 π n L/ λ = π or L= λ /(2 n) for smallest L.
17- (a) increases; (b) first order.
18- 111. The key trigonometric identity used in this proof is sin(2) = 2sin cos. Now,
we wish to show that Eq. 36-19 becomes (when d = a) the pattern for a single slit of
width 2a (see Eq. 36-5 and Eq. 36-6):
sin(2asin/)
I() = Im 
 .
 2asin/ 
2
We note from Eq. 36-20 and Eq. 36-21, that the parameters  and  are identical in this
case (when d = a), so that Eq. 36-19 becomes
cos(asin/)sin(asin/)
I() = Im 
 .
asin/


2
Multiplying numerator and denominator by 2 and using the trig identity mentioned
above, we obtain
2cos(asin/)sin(asin/)
sin(2asin/)
I() = Im 
 = Im 

2asin/


 2asin/ 
2
which is what we set out to show.
2