Light travels as transverse waves and faster than sound. It can be reflected, refracted and dispersed. Ray diagrams show what happens to light in mirrors and lenses. Eyes and cameras detect light.
Answer:
a

b

Explanation:
From the question we are told that
Their distance apart is 
The wavelength of each source wave 
Let the distance from source A where the construct interference occurred be z
Generally the path difference for constructive interference is

Now given that we are considering just the straight line (i.e points along the line connecting the two sources ) then the order of the maxima m = 0
so

=> 
=> 
Generally the path difference for destructive interference is

=> 
=> 
substituting values

=> 
So


and

=> 
=> 
A) Work energy relation;
Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
F = 900/0.04
= 22500 N
Therefore, force is 22500 N
b) From newton's second law of motion;
F = Ma
Thus; a = F/m
= 22500/(5×10^-3)
= 4,500,000 m/s²
But v = u-at
0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds
Answer:

Explanation:
P = Acoustic power = 63 µW
r = Distance to the sound source = 210 m
Acoustic power

Threshold intensity = 
Ratio

Ratio of the acoustic intensity produced by the juvenile howler to the reference intensity is 113.68
Let t = Theta and p = Phi
Tan t = y/x Then x =y/Tant.
Tant = y/(x-d) x-d = y/Tanp
y/Tant - d = y/Tanp
y -d*Tanr = y*Tant/Tanp
y-y*Tant/Tanp = d*Tanr
y(1 - Tanr/Tanp = d*Tant
y = d*Tant/(1-Tant/Tanp)