The wavelength of the light that are not reflected in the air is,
λ5= 600 nm, λ6= 500 nm, λ7=428 nm.
<u>Explanation:</u>
- For the above problem, we have the case as no<n1 and n2 < n1 .
- so we have(1<1.5;1.4<1.5)To find the wavelengths of the light that is not reflected in the air.
- Since the phase changes are not common on both the surfaces/
- we have the formula, λ=2d n²/m ( where m=1,2,3,4,5...)
- solving the problem we get,
- λ5= 600 nm, λ6= 500 nm, λ7=428 nm
- The wavelength of the light that are not reflected in the air is,
λ5= 600 nm, λ6= 500 nm, λ7=428 nm.
Answer: 100 J
Explanation: 1/2 5 x 2^2 = 100
Hope this made any sense.
External = R
Internal = r
Volume of hemisperical = 2/3 π(R³-r³)
V= 2/3 π(9.1³ - 8.4³)
V= 336.9 cm³
Answer:
a) 46.5º b) 64.4º
Explanation:
To solve this problem we will use the laws of geometric optics
a) For this part we will use the law of reflection that states that the reflected and incident angle are equal
θ = 43.5º
This angle measured from the surface is
θ_r = 90 -43.5
θ_s = 46.5º
b) In this part the law of refraction must be used
n₁ sin θ₁ = n₂. Sin θ₂
sin θ₂ = n₁ / n₂ sin θ₁
The index of air refraction is n₁ = 1
The angle is this equation is measured between the vertical line called normal, if the angles are measured with respect to the surface
θ_s = 90 - θ
θ_s = 90- 43.5
θ_s = 46.5º
sin θ₂ = 1 / 1.68 sin 46.5
sin θ₂ = 0.4318
θ₂ = 25.6º
The angle with respect to the surface is
θ₂_s = 90 - 25.6
θ₂_s = 64.4º
measured in the fourth quadrant
Explanation:
It is given that,
Magnetic field, B = 0.15 T
Charge on a proton, 
Mass of a proton, 
The cyclotron frequency is given by :
f = 2286785.40 Hz
or
Hence, this is the required solution.
