On a hot summer night, the refractive index of air is the smallest near the ground and increases with a height from the ground. When a light beam is directed horizontally, the Huygens' principle leads us to conclude that as it travels, the light beam,

We begin by recalling the Huygens-Fresnel principle. It tells us that every point of a wave-front is the source of secondary spherical wavelets, and the new wave-front at a later instant is the common tangent (envelope) to all those wavelets. In a medium where the speed of light changes from point to point, the radii of the successive secondary wavelets are different at different points, so the envelope is no longer a straight line and the ray (which is always drawn perpendicular to the wave-front) bends.

Let us translate the verbal description of the atmosphere given in the question into a refractive-index function:

Near the ground the refractive index is minimum and as we go to a height $$y$$ above the ground the refractive index $$n$$ increases. Symbolically, we may write

$$\frac{dn}{dy}>0.$$

The speed of light in any medium is related to the refractive index by the simple formula

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

where $$c$$ is the speed of light in vacuum. So if $$n$$ increases with $$y$$, the speed $$v$$ decreases with $$y$$, i.e.

$$\frac{dv}{dy}<0.$$

Now imagine the initial wave-front to be vertical because the beam is launched perfectly horizontal. Consider two neighbouring points on that wave-front: point $$A$$ at a small height $$y$$ and point $$B$$ infinitesimally lower, at height $$y-\Delta y$$. Because $$n(y)>n(y-\Delta y)$$, the speed at $$A$$, which is $$v(y)=c/n(y),$$ is smaller than the speed at $$B$$,

$$v(y)<v(y-\Delta y).$$

In an infinitesimal time interval $$\Delta t$$ the secondary wavelet from $$B$$ therefore advances a greater horizontal distance than the one from $$A$$. Geometrically the top portion of the wave-front lags behind the bottom portion, so the wave-front tilts such that its upper end stays behind. Because the ray is everywhere perpendicular to the wave-front, the ray tilts so that it starts pointing slightly upward.

We can make this more algebraic by invoking the differential form of Snell’s law for a stratified medium. Snell’s law in its usual form for two media is

$$n_1\sin\theta_1 = n_2\sin\theta_2.$$

For a medium whose refractive index varies continuously with height, the product $$n\sin\theta$$ remains constant along the ray. Hence, writing the constant as $$K$$, we have for every position $$y$$,

$$n(y)\sin\theta(y)=K.$$

Differentiating with respect to $$y$$,

$$\frac{d}{dy}\left[n(y)\sin\theta(y)\right]=0.$$

Using the product rule,

$$\sin\theta\,\frac{dn}{dy}+n\cos\theta\,\frac{d\theta}{dy}=0.$$

Solving for $$\dfrac{d\theta}{dy}$$ we get

$$\frac{d\theta}{dy}=-\frac{\sin\theta}{n\cos\theta}\,\frac{dn}{dy}.$$ $$\displaystyle\Longrightarrow\quad\frac{d\theta}{dy}=-\tan\theta\;\frac{1}{n}\,\frac{dn}{dy}.$$

We already established that $$\dfrac{dn}{dy}>0$$; $$n$$ itself is positive, and for a near-horizontal ray the initial $$\theta$$ (measured from the normal) is less than $$90^{\circ}$$, so $$\tan\theta>0$$. Therefore

$$\frac{d\theta}{dy}<0.$$

A negative derivative means that $$\theta$$ decreases as we move upward, i.e. the ray turns toward the higher layer (upward). So the continuously varying atmosphere behaves like a prism that bends the initially horizontal beam upward.

Hence, the Huygens’ principle together with the given variation of refractive index leads us to conclude that the light beam bends upward.

Hence, the correct answer is Option A.