IPhO 1993 · Theory problem 2

Laser Forces on a Transparent Prism

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General Tabulated Information

Quantity Symbol Value
Earth’s mean radius \(R_E\) \(6.4\times10^6\ \mathrm{m}\)
acceleration due to gravity \(g\) \(9.8\ \mathrm{m\,s^{-2}}\)
Newtonian gravitational constant \(G\) \(6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\)
permittivity of vacuum \(\varepsilon_0\) \(8.85\times10^{-12}\ \mathrm{C^2\,N^{-1}\,m^{-2}}\)
permeability of vacuum \(\mu_0\) \(8.85\times10^{-7}\ \mathrm{N\,A^{-2}}\) [sic; correct value \(4\pi\times10^{-7}\ \mathrm{N\,A^{-2}}\)]
speed of light in vacuum (or air) \(c\) \(3.00\times10^8\ \mathrm{m\,s^{-1}}\)
elementary charge \(e\) \(1.60\times10^{-19}\ \mathrm{C}\)
mass of electron \(m_e\) \(9.11\times10^{-31}\ \mathrm{kg}\)
mass of proton \(m_p\) \(1.67\times10^{-27}\ \mathrm{kg}\)
Planck constant \(h\) \(6.63\times10^{-34}\ \mathrm{J\,s}\)
Avogadro constant \(N_A\) \(6{,}02\times10^{23}\ \mathrm{mol^{-1}}\)
Boltzmann constant \(k\) \(1.38\times10^{-23}\ \mathrm{J\,K^{-1}}\)
molar gas constant \(R\) \(8.31\ \mathrm{J\,mol^{-1}\,K^{-1}}\)

LASER FORCES ON A TRANSPARENT PRISM

By means of refraction a strong laser beam can exert appreciable forces on small transparent objects. To see that this is so, consider a small glass triangular prism with an apex angle \(A=\pi-2\alpha\), a base of length \(2h\) and a width \(w\). The prism has an index of refraction \(n\) and a mass density \(\rho\).

Suppose that this prism is placed in a laser beam travelling horizontally in the \(x\) direction. (Throughout this problem assume that the prism does not rotate, i.e., its apex always points opposite to the direction of the laser beam, its triangular faces are parallel to the \(xy\) plane, and its base is parallel to the \(yz\) plane, as shown in Fig. 1.) Take the index of refraction of the surrounding air to be \(n_{\mathrm{air}}=1\). Assume that the faces of the prism are coated with an anti-reflection coating so that no reflection occurs.

Figure 1

Fig. 1.

The laser beam has an intensity that is uniform across its width in the \(z\) direction but falls off linearly with distance \(y\) from the \(x\) axis such that it has a maximum value of \(I_0\) at \(y=0\) and falls to zero at \(y=\pm4h\) (Fig. 2). [Intensity is power per unit area, e.g. expressed in \(\mathrm{W\,m^{-2}}\).]

Figure 2

Fig. 2.

1) Write equations from which the angle \(\theta\) (see Fig. 3) may be determined (in terms of \(\alpha\) and \(n\)) in the case when laser light strikes the upper face of the prism.

Figure 3

Fig. 3.

2) Express, in terms of \(I_0\), \(\theta\), \(h\), \(w\) and \(y_0\), the \(x\) and \(y\) components of the net force exerted on the prism by the laser light when the apex of the prism is displaced a distance \(y_0\) from the \(x\) axis where \(|y_0|\leq3h\).

Plot graphs of the values of the horizontal and vertical components of force as functions of vertical displacement \(y_0\).

3) Suppose that the laser beam is 1 mm wide in the \(z\) direction and \(80\,\mu\mathrm{m}\) thick (in the \(y\) direction). The prism has \(\alpha=30^\circ\), \(h=10\,\mu\mathrm{m}\), \(n=1.5\), \(w=1\,\mathrm{mm}\) and \(\rho=2.5\,\mathrm{g\,cm^{-3}}\).

How many watts of laser power would be required to balance this prism against the pull of gravity (in the \(-y\) direction) when the apex of the prism is at a distance \(y_0=-h/2\) (\(=-5\,\mu\mathrm{m}\)) below the axis of the laser beam?

4) Suppose this experiment is done in the absence of gravity with the same prism and a laser beam with the same dimensions as in (3), but with \(I_0=10^8\ \mathrm{W\,m^{-2}}\).

What would be the period of the oscillations that occur when the prism is displaced and released a distance \(y=h/20\) from the center line of the laser beam?

Official English text of the problem.

Answers

  1. If \(\beta\) is the angle of refraction at the upper face and \(\gamma\) is the incidence angle at the base, \[ \boxed{\sin\alpha=n\sin\beta,\qquad \gamma=\alpha-\beta,\qquad \sin\theta=n\sin\gamma.} \] Equivalently, \[ \boxed{\theta=\arcsin\!\left[ \sin\alpha\left(\sqrt{n^2-\sin^2\alpha}-\cos\alpha\right) \right].} \]

  2. Define \[ u=\frac{y_0}{h},\qquad C_x=\frac{I_0wh}{4c}(1-\cos\theta),\qquad C_y=\frac{I_0wh}{4c}\sin\theta. \] Then \[ \boxed{ F_x=C_x \begin{cases} 7-u^2,& |u|\leq1,\\ 8-2|u|,&1\leq|u|\leq3, \end{cases}} \] \[ \boxed{ F_y=-C_y \begin{cases} u(2-|u|),&|u|\leq1,\\ \operatorname{sgn}(u),&1\leq|u|\leq3. \end{cases}} \] The horizontal-force graph is even, with a central parabolic maximum and straight outer segments. The vertical-force graph is odd and restoring, with central quadratic segments and constant outer segments.

  3. The required total laser power is \[ \boxed{P\simeq33\ \mathrm{W}.} \] The corresponding peak intensity is \(I_0\simeq8.3\times10^8\ \mathrm{W\,m^{-2}}\).

  4. The small-oscillation approximation gives \[ \boxed{T_0=2\pi\sqrt{\frac{2\rho c h^2\tan\alpha}{I_0\sin\theta}} \simeq1.12\times10^{-2}\ \mathrm{s}.} \] Retaining the weak nonlinearity at the specified amplitude \(h/20\) gives \[ \boxed{T\simeq1.13\times10^{-2}\ \mathrm{s},} \] about \(1.08\%\) longer.

Solution

We use geometrical optics, neglect absorption as appropriate for the transparent prism, and apply the stipulated absence of reflection. Forces are calculated to leading order in the prism’s speed divided by \(c\).

1) Refraction and the outgoing angle

The upper face makes an angle \(\alpha\) with the vertical base. Its inward normal therefore points at an angle \(\alpha\) below the horizontal. The horizontal incident ray has incidence angle \(\alpha\).

Let \(\beta\) be the refracted angle measured from this normal. Snell’s law gives \[ \sin\alpha=n\sin\beta. \] The ray inside the prism consequently travels at an angle \[ \gamma=\alpha-\beta \] below the horizontal.

The normal to the exit face is horizontal, so \(\gamma\) is also the incidence angle there. Applying Snell’s law again, \[ n\sin\gamma=\sin\theta. \] Thus \[ \theta=\arcsin\!\left[ n\sin\left(\alpha-\arcsin\frac{\sin\alpha}{n}\right) \right]. \]

An equivalent expression follows by expanding the sine: \[ \begin{aligned} \sin\theta &=n(\sin\alpha\cos\beta-\cos\alpha\sin\beta)\\ &=\sin\alpha\sqrt{n^2-\sin^2\alpha} -\sin\alpha\cos\alpha. \end{aligned} \] Hence \[ \boxed{\sin\theta= \sin\alpha\left(\sqrt{n^2-\sin^2\alpha}-\cos\alpha\right).} \]

The transmitted path shown in Fig. 3 requires \(n\sin\gamma\leq1\), which is satisfied by the numerical data in the problem. By symmetry, light entering the lower face exits at angle \(\theta\) above the horizontal.

2) Force components and graphs

The intensity profile is \[ I(y)=I_0\left(1-\frac{|y|}{4h}\right), \qquad |y|\leq4h. \]

Let \(P_+\) and \(P_-\) be the powers entering the upper and lower faces. When the apex is at \(y_0\), their projected intervals perpendicular to the incident beam are respectively \[ [y_0,y_0+h],\qquad [y_0-h,y_0]. \] Therefore \[ P_+=w\int_{y_0}^{y_0+h}I(y)\,dy,\qquad P_-=w\int_{y_0-h}^{y_0}I(y)\,dy. \] The area element here is \(w\,dy\), the area projected perpendicular to the incident light. No additional incidence-angle factor is needed.

Light of power \(P\) in air carries momentum at a rate \(P/c\). For light entering the upper face, the incident and outgoing directions are \[ (1,0),\qquad(\cos\theta,-\sin\theta). \] The force on the prism equals the incoming momentum flux minus the outgoing momentum flux: \[ \mathbf F_+=\frac{P_+}{c} (1-\cos\theta,\ \sin\theta). \] Likewise, \[ \mathbf F_-=\frac{P_-}{c} (1-\cos\theta,\ -\sin\theta). \] Adding, \[ \boxed{ F_x=\frac{1-\cos\theta}{c}(P_++P_-),\qquad F_y=\frac{\sin\theta}{c}(P_+-P_-). } \]

To evaluate the powers, put \(u=y_0/h\). An antiderivative of the dimensionless intensity profile is \[ J(v)=v-\frac{v|v|}{8}, \qquad J'(v)=1-\frac{|v|}{4}. \] Consequently, \[ \frac{P_+}{I_0wh}=J(u+1)-J(u),\qquad \frac{P_-}{I_0wh}=J(u)-J(u-1). \]

For \(|u|\leq1\), the prism straddles the beam axis. Evaluating these expressions gives \[ P_++P_-=\frac{I_0wh}{4}(7-u^2), \] \[ P_+-P_-=-\frac{I_0wh}{4}u(2-|u|). \]

For \(1\leq|u|\leq3\), the entire prism lies on one side of the axis, where intensity varies linearly: \[ P_++P_-=\frac{I_0wh}{4}(8-2|u|), \] \[ P_+-P_-=-\frac{I_0wh}{4}\operatorname{sgn}(u). \]

Thus, with \[ C_x=\frac{I_0wh}{4c}(1-\cos\theta),\qquad C_y=\frac{I_0wh}{4c}\sin\theta, \] the requested forces are \[ \boxed{ F_x=C_x \begin{cases} 7-u^2,&|u|\leq1,\\ 8-2|u|,&1\leq|u|\leq3, \end{cases} \qquad F_y=-C_y \begin{cases} u(2-|u|),&|u|\leq1,\\ \operatorname{sgn}(u),&1\leq|u|\leq3. \end{cases}} \]

The graph values are:

\(u=y_0/h\) \(F_x/C_x\) \(F_y/C_y\)
\(-3\) \(2\) \(1\)
\(-2\) \(4\) \(1\)
\(-1\) \(6\) \(1\)
\(-1/2\) \(6.75\) \(0.75\)
\(0\) \(7\) \(0\)
\(1/2\) \(6.75\) \(-0.75\)
\(1\) \(6\) \(-1\)
\(2\) \(4\) \(-1\)
\(3\) \(2\) \(-1\)

For the horizontal-force graph, draw a downward-opening parabola between \(u=-1\) and \(u=1\), with maximum \(7C_x\) at the origin. Continue with straight segments to \(2C_x\) at \(u=\pm3\). The force remains positive throughout the specified interval.

For the vertical-force graph, draw the constant value \(+C_y\) from \(u=-3\) to \(-1\), then the quadratic segment decreasing to zero at \(u=0\). Reflect this segment through the origin to reach \(-C_y\) at \(u=1\), then continue horizontally to \(u=3\). Its slope at the origin is \(-2C_y/h\). Both force graphs join with continuous slopes at \(u=\pm1\).

The vertical force is restoring: a prism below the axis receives more light on its upper face and is pushed upward; above the axis it is pushed downward.

3) Laser power required to balance gravity

The distance from the apex to the base is \[ d=h\tan\alpha. \] The volume and mass are therefore \[ V=\frac12(2h)d\,w=h^2w\tan\alpha, \qquad m=\rho h^2w\tan\alpha. \] Using SI units, \[ h=10^{-5}\ \mathrm m,\qquad w=10^{-3}\ \mathrm m,\qquad \rho=2.5\times10^3\ \mathrm{kg\,m^{-3}}, \] so \[ m=1.4434\times10^{-10}\ \mathrm{kg}, \qquad mg=1.4145\times10^{-9}\ \mathrm N. \]

For \(\alpha=30^\circ\) and \(n=1.5\), \[ \beta=\arcsin\left(\frac{\sin30^\circ}{1.5}\right) =19.471^\circ, \] \[ \gamma=10.529^\circ,\qquad \theta=15.908^\circ,\qquad \sin\theta=0.274094. \]

At \(y_0=-h/2\), we have \(u=-1/2\), giving \[ F_y=\frac34C_y =\frac{3I_0wh\sin\theta}{16c}. \] Vertical balance requires \[ \frac{3I_0wh\sin\theta}{16c} =\rho h^2w\tan\alpha\,g. \] Hence \[ I_0=\frac{16\rho ghc\tan\alpha}{3\sin\theta} =8.2571\times10^8\ \mathrm{W\,m^{-2}}. \]

The required power is the power of the whole beam. Its thickness \(80\,\mu\mathrm m\) equals \(8h\), and the area under its triangular intensity profile is \(4hI_0\). Thus \[ P=w\int_{-4h}^{4h}I(y)\,dy=4hwI_0. \] Substitution gives \[ P=4(10^{-5})(10^{-3})(8.2571\times10^8) =33.03\ \mathrm W. \] Therefore \[ \boxed{P\simeq33\ \mathrm W.} \]

This balances gravity in the \(y\) direction, as requested. The horizontal optical force remains positive.

4) Oscillation period without gravity

Throughout the stated oscillation, \(|y_0|\leq h/20<h\). The central force expression can be written \[ F_y=-\frac{I_0w\sin\theta}{2c}\, y_0\left(1-\frac{|y_0|}{2h}\right). \] Define \[ K=\frac{I_0w\sin\theta}{2c}. \] At amplitude \(a=h/20\), the nonlinear factor differs from unity by at most \[ \frac{a}{2h}=\frac1{40}=2.5\%. \] The leading approximation is therefore simple harmonic motion: \[ m\ddot y_0=-Ky_0,\qquad \omega_0^2=\frac Km. \] Using the mass found above, \[ \omega_0^2 =\frac{I_0\sin\theta}{2\rho c h^2\tan\alpha}, \] and \[ \boxed{ T_0=2\pi\sqrt{\frac{2\rho c h^2\tan\alpha}{I_0\sin\theta}}. } \]

For \(I_0=10^8\ \mathrm{W\,m^{-2}}\), \[ K=4.5682\times10^{-5}\ \mathrm{N\,m^{-1}}, \qquad \omega_0=562.58\ \mathrm{s^{-1}}, \] so \[ \boxed{T_0=0.0111685\ \mathrm s\approx11.2\ \mathrm{ms}.} \]

The finite-amplitude correction can also be evaluated from the exact force. Taking the potential energy to vanish at the origin, \[ U(y_0)=\frac K2y_0^2-\frac{K}{6h}|y_0|^3. \] Release from rest at \(y_0=a\) gives \[ \frac12m\dot y_0^2=U(a)-U(y_0). \] By symmetry, the period is four times the travel time from \(0\) to \(a\): \[ T=4\sqrt{\frac mK} \int_0^a \frac{dy}{\sqrt{a^2-y^2-(a^3-y^3)/(3h)}}. \] Setting \(s=y/a\), \[ T=\frac4{\omega_0} \int_0^1 \frac{ds}{\sqrt{1-s^2-\dfrac{a}{3h}(1-s^3)}}. \] For \(a/h=1/20\), numerical evaluation yields \[ \frac{T}{T_0}=1.01079, \] and hence \[ \boxed{T=0.0112890\ \mathrm s\approx11.3\ \mathrm{ms}.} \] The period is therefore about \(0.011\ \mathrm s\); the small-oscillation approximation is accurate to approximately \(1.1\%\) at the specified amplitude.

The solution was checked against the official solution.