1. System Setup and Coordinates

Let the centre of the globe be the origin $(0,0)$. The radius of the globe is $R = d/2$. The spider is located at position $\vec{S} = \frac{d}{\sqrt{3}} \hat{j}$ (along the vertical north-pole axis). The refractive index of the globe is $\mu = 2$.

2. Determining the Horizon of the Spider

Due to the reversibility of light, the spider can see any point on the globe if a light ray from that point can reach the spider. The limit of the spider’s direct vision is defined by the tangent cone from the spider to the sphere.

Consider a triangle formed by the Origin $O$, the Spider $S$, and the tangent point $P$. $$ OS = \frac{d}{\sqrt{3}} = \frac{2R}{\sqrt{3}} $$ $$ OP = R $$ In the right-angled triangle $\Delta OPS$ (right angled at tangent point P? No, tangent is perpendicular to radius): $$ \cos \theta = \frac{OP}{OS} = \frac{R}{2R/\sqrt{3}} = \frac{\sqrt{3}}{2} $$ $$ \theta = 30^\circ $$ This polar angle $\theta = 30^\circ$ corresponds to a latitude of $90^\circ – 30^\circ = 60^\circ$ North. Thus, the spider can directly see the cap from the North Pole down to $60^\circ$ N.

3. Refraction at the Grazing Limit

Light incident at this limiting angle (grazing incidence, $i = 90^\circ$) will enter the sphere and refract. Using Snell’s Law at point $P$: $$ 1 \cdot \sin(90^\circ) = \mu \sin(r) $$ $$ 1 = 2 \sin(r) \implies \sin(r) = \frac{1}{2} \implies r = 30^\circ $$

At point $P$ (polar angle $30^\circ$), the normal makes an angle of $30^\circ$ with the vertical axis $\hat{j}$. Since the angle of refraction $r$ is also $30^\circ$ “inwards” from the normal, the refracted ray travels parallel to the vertical axis (straight down).

4. Mapping the Visible Region

The refracted rays from the “horizon” ($60^\circ$ N) travel vertically downwards through the sphere. A vertical ray starting at $x = R \sin(30^\circ) = R/2$ will strike the bottom surface at $x = R/2$. To find the corresponding latitude on the bottom hemisphere: The coordinate $x = R/2$ on the sphere corresponds to a polar angle $\phi$ (measured from the South Pole) where: $$ \sin \phi = \frac{x}{R} = \frac{R/2}{R} = \frac{1}{2} \implies \phi = 30^\circ $$ This corresponds to a latitude of $60^\circ$ South.

Therefore, the spider can see: 1. The top cap directly ($90^\circ$ N to $60^\circ$ N). 2. The bottom cap via refraction through the sphere ($90^\circ$ S to $60^\circ$ S).

The region that remains hidden (cannot be seen) is the band between these limits.

Conclusion

The fly cannot be seen if it sits between latitudes $60^\circ$ N and $60^\circ$ S.