Month: February 2024

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202402071059 Solution to 1980-CE-PHY-II-16

A light ray passes through a spherical air bubble in water. Which of the following represents the path of the emergent ray?

A. P
B. Q
C. R
D. S

Roughwork.

Refractive (/refraction) index of a pure substance, with some exceptions, is proportional to its density; and of a mixture of substances, as a rule, to its mass concentration.

Thus water, higher in density than air, should have a larger refractive index than air:

n_{\textrm{water}}>n_{\textrm{air}};

alternatively, as light travels in air faster than in water, from

\displaystyle{v=\frac{c}{n}},

it can be seen also.

Let \theta_{1} and \theta_{2} be the angle of incidence and of refraction when light is refracted through the first interface; and \theta_{2} and \theta_{3}, when through the second interface.

Assuming WLOG n_1\geqslant n_2 and noting \sin \theta \geqslant 0 (non-negative) for \theta\in \big[ 0,\frac{\pi}{2}\big]. By Snell’s Law, write

\begin{aligned} n_1\sin\theta_1 & = n_2\sin\theta_2 \\ \frac{\sin\theta_1}{\sin\theta_2} & = \frac{n_2}{n_1}\leqslant 1 \\ \sin\theta_1 & \leqslant \sin\theta_2 \\ \theta_1 & \leqslant \theta_2 \\ \end{aligned}

\therefore \quad n_1\geqslant n_2\Longrightarrow \theta_2\geqslant \theta_1

\therefore\quad n_{\textrm{water}}>n_{\textrm{air}}\Longrightarrow \theta_{\textrm{air}}>\theta_{\textrm{water}}

Answer. Path P.

Reverse the direction of incidence and emergence:

Light ray along path: P bends away from and then towards the normal; Q bends towards the normal once and again; R bends away from the normal once and again; S bends towards and then away from the normal. Wits have it that Q and R are unlikely. So we turn our attention to light rays P and S.

Fussy enough, path P is symmetric with respect to straight line L_P; but there across path S is no line of symmetry were it to be L_S.

This problem is not to be attempted.

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202402051319 Pastime Exercise 009

The blogger claims no originality of his problem below.

On a rollover road,

a vehicle performs circular motion:

Discuss how the driver could prevent a traffic accident.

Roughwork.

Take positive both, the angle \theta in an anti-clockwise direction, and the displacement \mathbf{s} in a rightward and an upward direction.

\begin{aligned} \mathbf{W} & = m\mathbf{g} = -mg\,\hat{\mathbf{j}} \\ \mathbf{N}(\theta ) & = -N(\theta )\,\hat{\mathbf{r}} \\ \mathbf{N}(\theta ) & = N_x(\theta )\,\hat{\mathbf{i}} + N_y(\theta )\,\hat{\mathbf{j}} \\ N^2(\theta ) & = N_x^2(\theta )+N_y^2(\theta ) \\ N_x(\theta ) & = N\sin\theta \\ N_y(\theta ) & = N\cos\theta \\ \mathbf{f}(\theta ) & = -\mu N(\theta )\,\hat{\boldsymbol{\theta}} \\ \mathbf{f}(\theta ) & \perp \mathbf{N}(\theta ) \\ \end{aligned}

In order for the vehicle not to leave the track, the radius of curvature r \textrm{\scriptsize{MUST}} be kept constant, i.e., r=R(=\textrm{Const.}).

Recall, in uniform circular motion there isn’t any (angular) acceleration in angular velocity, i.e.,

\begin{aligned} \alpha (t) & = 0 \\ \frac{\mathrm{d}}{\mathrm{d}t}\big(\omega (t)\big) & = 0 \\ \frac{\mathrm{d}}{\mathrm{d}t}\bigg( \frac{\mathrm{d}}{\mathrm{d}t}\big(\theta (t)\big)\bigg) & = 0 \\ \frac{\mathrm{d}}{\mathrm{d}t}\big(\theta (t)\big) & = a \\ \theta (t) & = at+b\qquad (\exists\, a,b\in\mathbb{R}) \\ \end{aligned}

If uniform circular motion is \textrm{\scriptsize{NOT}} assumed,

i.e., \alpha \neq 0\textrm{ \scriptsize{OR} }r\neq\textrm{Const.},

then angular displacement \theta (t) wrt time t will be a polynomial in an indeterminate t of degree 2 or higher,

i.e., O(t)=O(t^c)\qquad\exists\, c\, (\in\mathbb{Z}^+)\geqslant 2.

Angular acceleration \boldsymbol{\alpha} is less often mentioned than is its parallel tangential acceleration \mathbf{a}_{\parallel}\, (=r\boldsymbol{\alpha}) as the pair to centripetal (/centrifugal) acceleration \mathbf{a}_{\perp}.

By Newton 2nd Law, write

\begin{aligned} \mathbf{F}_{\textrm{net}} & = m\mathbf{a} \\ \mathbf{W}+\mathbf{N}+\mathbf{f} & = m(\mathbf{a}_{\parallel}+\mathbf{a}_{\perp}) \\ \end{aligned}

The presence of a centripetal acceleration, i.e., \exists\,\mathbf{a}_{\perp}\neq\mathbf{0}, is necessary for any circular motion, be it uniform or not; but not sufficient for the absence of a tangential acceleration \forall\,\mathbf{a}_{\parallel}= 0 (why?). One knows instinctively, that the vehicle should possess a minimum angular speed \min (\omega ) to do this dangerous stunt. And so much the better if beyond this bound \min (\omega )\leqslant \omega \leqslant \max (\omega ) \ll c one has all degrees of freedom.

By definition,

\begin{aligned} \mathbf{a}_{\parallel} & = R\dot{\omega}\,\hat{\boldsymbol{\theta}} = R\ddot{\theta}\,\hat{\boldsymbol{\theta}} \\ \mathbf{a}_{\perp} & = -R\omega^2\,\hat{\mathbf{r}} = -R\dot{\theta}^2\,\hat{\mathbf{r}} \\ \end{aligned}

Stepping on and off the accelerator (/gas pedal), however delicately, is \textrm{\scriptsize{NOT}} designed to keep a constant acceleration for the gear.

This problem is not to be attempted.