(a) A homogeneous right triangular pyramid with the base side a and height 3a/2 is shown below. Obtain the moment of inertia…

(a) The standard moment of inertia tensor is about the centre of mass. From the figure the tetrahedron has vertices O(0,0,0), A(a,0,0), B(0,a,0), C(0,0,3a/2); put h=3a/2. Its volume is V=a²h/6=a³/4, so for mass M the density is ρ=4M/a³. The centroid of a tetrahedron is the average of its vertices, r₀=(O+A+B+C)/4=(a/4,a/4,3a/8). The same simplex formula with p=1 and q=r=0 gives ∫x dV=aV/4, and similarly for y and z, confirming the centroid.

For the simplex x,y,z≥0, x/a+y/a+z/h≤1, set X=x/a, Y=y/a, Z=z/h. Dirichlet's simplex integral, valid for non-negative integers p,q,r, is obtained by integrating Z first over 0 to 1-X-Y: ∫ x^p y^q z^r dV = a^(p+1) a^(q+1) h^(r+1) times the integral over the unit simplex of X^p Y^q Z^r dX dY dZ. The Z integral gives (1-X-Y)^(r+1)/(r+1). Then ∫ from 0 to 1-X of Y^q (1-X-Y)^(r+1) dY = (1-X)^(q+r+2) q!(r+1)!/(q+r+2)!, and ∫ from 0 to 1 of X^p (1-X)^(q+r+2) dX = p!(q+r+2)!/(p+q+r+3)!. Multiplying by 1/(r+1) gives p!q!r!/(p+q+r+3)!. The hypotheses are met here because the pyramid is exactly this simplex and the required exponents are 0, 1 and 2. Hence ∫x²dV=∫y²dV=a⁴h/60=a⁵/40, ∫z²dV=a²h³/60=9a⁵/160, ∫xy dV=a⁴h/120=a⁵/80, ∫xz dV=∫yz dV=a³h²/120=3a⁵/160.

About O, using Ixx=∫(y²+z²)dm and Ixy=-∫xy dm: Ixx(O)=Iyy(O)=ρ(∫y²+∫z²)=13Ma²/40, Izz(O)=ρ(∫x²+∫y²)=Ma²/5, Ixy(O)=-ρ∫xy=-Ma²/20, Ixz(O)=Iyz(O)=-ρ∫xz=-3Ma²/40.

The tensor parallel-axis theorem is I(O)=I(CM)+M[(r₀·r₀)1-r₀r₀ᵀ], so I(CM)=I(O)-M[(r₀·r₀)1-r₀r₀ᵀ]. Here r₀·r₀=17a²/64. Componentwise, Ixx(CM)=Ixx(O)-M(17a²/64-a²/16)=Ma²(13/40-13/64)=39Ma²/320, Iyy(CM)=39Ma²/320, Izz(CM)=Izz(O)-M(17a²/64-9a²/64)=Ma²(1/5-1/8)=3Ma²/40, Ixy(CM)=Ixy(O)+M(a/4)(a/4)=Ma²(-1/20+1/16)=Ma²/80, Ixz(CM)=Iyz(CM)=Ixz(O)+M(a/4)(3a/8)=Ma²(-3/40+3/32)=3Ma²/160.

I(CM)=Ma²[[39/320,1/80,3/160],[1/80,39/320,3/160],[3/160,3/160,3/40]], with units kg m² if M is in kg and a in m. As a check, the trace about O is 17Ma²/20 and the trace about CM is 51Ma²/160; their difference is 17Ma²/32=2M(r₀·r₀), as the parallel-axis theorem requires. The matrix is symmetric and positive definite; the off-diagonal magnitudes are small compared with the diagonal entries.

(b) The air film has refractive index μ=1. For paraxial rays the film thickness is t=t₀+r²/(2R), where t₀ is any small central thickness or counting offset. The sagitta of a sphere is t=R-sqrt(R²-r²); for r≪R this becomes r²/(2R), and here the largest radius is below 0.44 cm while R=100 cm, so the approximation is excellent. In reflected light one reflection has a phase reversal, so the nth bright ring, with n=1 for the first bright ring, satisfies 2μ(t₀+r²/(2R))=(n-1/2)λ. Therefore Dₙ²=4r²=4(n-1/2)λR/μ+C, where C is a constant. C absorbs any central gap, dust, or mislabelling of the first bright ring; it is not needed because two diameters are given. If the ideal contact formula without C were forced through both given diameters, it would give different wavelengths, about 704 nm from D₄ and 568 nm from D₁₅; hence the constant term is essential for this data. The important point is that Dₙ² is linear in n with slope 4λR/μ; the same slope is obtained for transmitted bright rings, so the difference method is robust.

Given D₄=0.314 cm and D₁₅=0.574 cm, D₄²=0.098596 cm², D₁₅²=0.329476 cm². Thus s=(D₁₅²-D₄²)/(15-4)=0.230880/11=0.0209891 cm² per ring. Since s=4λR/μ and μ=1, R=100 cm, λ=s/(4R)=0.0209891/400=5.2473×10⁻⁵ cm=524.7 nm. Then D₂₄²=D₄²+(24-4)s=0.098596+20(0.0209891)=0.518378 cm², so D₂₄=0.720 cm. D₃₆²=D₄²+(36-4)s=0.098596+32(0.0209891)=0.770247 cm², so D₃₆=0.878 cm. Using D₁₅ instead of D₄ gives the same values: D₂₄²=0.329476+9s and D₃₆²=0.329476+21s. λ=5.247×10⁻⁵ cm (524.7 nm).

(c) In a He-Ne laser the He gas is not the main lasing species; it is the collisional pump medium. The electrical discharge excites He atoms efficiently because He has favourable excitation cross-sections and long-lived metastable states. The 2¹S₀ and 2³S₁ states are metastable because the one-photon electric-dipole transitions to the ground state are forbidden or highly forbidden, so they survive long enough for collisions. The gas is usually a He-Ne mixture with high He pressure and low Ne pressure, so He metastables are abundant and Ne levels are not strongly quenched. These metastable He atoms collide with Ne atoms and transfer their energy resonantly to the upper Ne laser levels. Ne has closely spaced levels, and the 5s and 3s levels lie almost exactly at the He metastable energies, making transfer efficient. The lower Ne levels are not directly pumped by the discharge and are rapidly depopulated by collisions and radiative transitions, so the upper level remains more populated than the lower level and stimulated emission gives net gain. Thus He converts electrical energy into metastable atomic energy and then into a Ne population inversion.

Energy ↑

• 20.6 eV: Ne 5s ← He 2¹S₀ (20.62 eV)
• 19.8 eV: Ne 3s ← He 2³S₁ (19.82 eV)
• 18.7 eV: Ne 3p (lower laser level)
• 16.8 eV: Ne 2p
• 0 eV: Ne ground, He ground

Collisional transfer: He* + Ne → He + Ne*. Laser transition: Ne 5s ↓ Ne 3p, 632.8 nm. The near-resonant transfers are He 2¹S₀ + Ne → He + Ne 5s and He 2³S₁ + Ne → He + Ne 3s. After the 5s→3p laser transition, Ne 3p decays to 2p and then to ground, so the lower level does not accumulate population. Because the 5s level is populated faster than the 3p level, the 5s→3p transition is the principal visible He-Ne laser line.