Atechnician places a fluid substance, with a refractive index of 1.61, between two horizontal panes of flat glass (each of which has n = 1.45). she then illuminates the assembly from above with monochromatic light (λ = 330 nm). what is the minimum thickness (in nm) that the liquid layer must have, if the light is to be strongly reflected back toward its source?

the minimum thickness (in nm)  t = 102.5 nm

Explanation:

t = thickness of the film

n = index of refraction = 1.61

m = order = 1

∝ = wavelength = 330 x 10-9 m

so we have the formula

2nt = m∝

2 (1.61) t = (1) (330 x 10-9)

then...

t = 1.025 x 10-7 m

converting...

t = 102.5 nm

trig 15.3 cos

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s