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ANTONII [103]
3 years ago
8

A 360-nm thick oil film floats on the surface of a pool of water. the indices of refraction of the oil and the water are 1.5 and

1.33, respectively. when the surface of the oil is illuminated from above at normal incidence with white light, what are the two wavelengths of light between 400 nm to 800 nm wavelength that are most strongly reflected?a) 410 nm and 700 nmb) 430 nm and 720 nmc) 450 nm and 740 nmd) 470 nm and 760 nme) 490 nm and 780 nm
Physics
1 answer:
jenyasd209 [6]3 years ago
7 0
The correct answer is 432, and 720.
The thickness of a film is t= 360nm
the refractive index of oil n₀t = (m +1/2) λ
For m =0
λ = 4n₀t
= 4(1.50)(360)
= 2160nm
for m = 1
λ = 4n₀t
= 4(1.50)(360)/3
= 720nm
m = 2
λ = 4n₀t/5 = 4(1.50)(360)/5
= 432nm
The wavelength which are most strongly reflected are
432nm, 720nm.
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Will kilogram of hydrogen contains more atoms than kilogram of lead?
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3 years ago
A 6.72 particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive direction,
Romashka [77]

Answer:

                    v_{y}  = -104 m/s

Explanation:

Using:

Force = electric field * charge

F=e*q

Force = magnitude of charge * velocity * magnetic field * sin tither

F_{x2}= |q|*v*B*sin \alpha

Force on particle due to electric field:

     F_{x1}= E*q = (1270N/C)*(-6.72*10^{-6} ) = -8.53*10^{-3}

Force on particle due to magnetic field:

F_{x2}= |q|*v*B*sin \alpha  = (6.72*10^{-6} )*(1.15)*(sin90)*v = (7.728*10^{-6})*(v)

F_{x2} is in the positive x direction as F_{x1} is in the negative x direction while net force is in the positive x direction.

Magnetic field is in the positive Z direction, net force is in the positive x direction.

According to right hand rule, Force acting on particle is perpendicular to the direction of magnetic field and velocity of particle. This would mean the force is along the y-axis. As this is a negatively charged particle, the direction of the velocity of the particle is reversed. Therefore velocity of particle, v, has to be in the negative y direction.

Now,

                    F_{xnet}- F_{x1 } = F_{x2 }

                    (6.13*10^{-3}) - (8.53*10^{-3} ) = (7.728*10^{-6})*(v)

                    v = (F_{xnet}  - F_{x1}) / (F_{x2} )

                        =((6.13*10^{-3} ) - (8.53*10^{-3})) / (7.728*10^{-6})

                       = (- 104.25) m/s

                      v_{y}  = -104 m/s

8 0
3 years ago
an object 50 cm high is placed 1 m in front of a converging lens whose focal length is 1.5 m. determine the image height (in cm)
Juli2301 [7.4K]

Given :

An object 50 cm high is placed 1 m in front of a converging lens whose focal length is 1.5 m.

To Find :

the image height (in cm).

Solution :

By lens formula :

\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}

Here, u = - 100 cm

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\dfrac{1}{v} - \dfrac{1}{-100} = \dfrac{1}{150}\\\\v = - 300 \ cm

Now, magnification is given by :

m = \dfrac{v}{u} = \dfrac{h_i}{h_o}\\\\h_i = \dfrac{300}{100}\times 1\\\\h_i = 3 \ m

Therefore, the image height is 3 m or 300 cm.

5 0
2 years ago
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