Question

A source of light is placed at S inside a glass of refractive index 2. P is center of sphere of radius R where PS = R/2. Then illuminated area of right hemisphere seen by the outside observer is of ______ percent of total area of sphere

Answer 1

25

Answer 2

To determine the illuminated area of the right hemisphere seen by an outside observer, we need to consider the conditions for light to emerge from the glass sphere into the air. This involves the concept of Total Internal Reflection (TIR).

1. Critical Angle (C): The light travels from glass (denser medium, refractive index $\mu_1 = 2$) to air (rarer medium, refractive index $\mu_2 = 1$). The critical angle $C$ is given by: $\sin C = \frac{\mu_2}{\mu_1} = \frac{1}{2}$ Therefore, $C = 30^\circ$. For light to emerge from the sphere, the angle of incidence ($i$) at the glass-air interface must be less than or equal to the critical angle ($i \le C$).

2. Geometry of the Ray: Let P be the center of the sphere and S be the position of the light source. We are given $PS = R/2$, where R is the radius of the sphere. Let Q be a point on the surface of the sphere where a light ray from S strikes. The line PQ is the normal to the surface at Q. The angle of incidence is $i = \angle SQP$. Let $\phi = \angle SPQ$. This angle $\phi$ defines the position of Q on the sphere relative to the line PS.

   Consider the triangle $\triangle SPQ$. The sides are $PS = R/2$, $PQ = R$, and $SQ$. Using the Law of Sines in $\triangle SPQ$: $\frac{PS}{\sin i} = \frac{PQ}{\sin(\angle QSP)}$ $\frac{R/2}{\sin i} = \frac{R}{\sin(\angle QSP)}$ This implies: $\sin(\angle QSP) = 2 \sin i$

3. Condition for Emergence: For light to emerge, $i \le C = 30^\circ$. Substituting this into the relation above: $\sin(\angle QSP) = 2 \sin i \le 2 \sin 30^\circ = 2 \times \frac{1}{2} = 1$ So, $\sin(\angle QSP) \le 1$. The maximum value of $\sin(\angle QSP)$ is 1, which occurs when $\angle QSP = 90^\circ$. This corresponds to the limiting case where $i = C = 30^\circ$.

4. Determining the Illuminated Region: In the limiting triangle $\triangle SPQ_{crit}$, where $i = 30^\circ$ and $\angle Q_{crit}SP = 90^\circ$: The sum of angles in a triangle is $180^\circ$. $\angle SPQ_{crit} + \angle Q_{crit}SP + \angle SQ_{crit}P = 180^\circ$ $\angle SPQ_{crit} + 90^\circ + 30^\circ = 180^\circ$ $\angle SPQ_{crit} = 180^\circ - 120^\circ = 60^\circ$.

   This means that any ray from S that strikes the sphere at a point Q such that $\angle SPQ \le 60^\circ$ will emerge. The illuminated region on the sphere is a spherical cap defined by this angle $\phi_{max} = 60^\circ$. The axis of this spherical cap is the line passing through P and S.

5. Area of the Illuminated Cap: The area of a spherical cap is given by the formula $A_{cap} = 2 \pi R^2 (1 - \cos \phi_{max})$, where $\phi_{max}$ is the half-angle of the cone from the center of the sphere. Substituting $\phi_{max} = 60^\circ$: $A_{cap} = 2 \pi R^2 (1 - \cos 60^\circ) = 2 \pi R^2 \left(1 - \frac{1}{2}\right) = 2 \pi R^2 \left(\frac{1}{2}\right) = \pi R^2$

6. Illuminated Area of the Right Hemisphere: Let P be at the origin (0,0,0). Let the line PS be along the positive x-axis, so S is at (R/2, 0, 0). The "right hemisphere" refers to the part of the sphere where $x \ge 0$. The spherical cap is centered on the positive x-axis (the point (R,0,0) on the sphere). For any point Q on this cap, its x-coordinate is $R \cos \phi$. Since $0 \le \phi \le 60^\circ$, we have $1/2 \le \cos \phi \le 1$. Thus, the x-coordinates of the points on the cap range from $R/2$ to $R$. All these points have $x \ge R/2$, which means they are entirely within the right hemisphere ($x \ge 0$). Therefore, the illuminated area of the right hemisphere is $\pi R^2$.

7. Percentage of Total Area of Sphere: The total surface area of the sphere is $A_{sphere} = 4 \pi R^2$. The question asks for the illuminated area of the right hemisphere as a percentage of the total area of the sphere. $\text{Percentage} = \frac{\text{Illuminated Area}}{\text{Total Area of Sphere}} \times 100\% = \frac{\pi R^2}{4 \pi R^2} \times 100\% = \frac{1}{4} \times 100\% = 25\%$

The illuminated area of the right hemisphere seen by the outside observer is 25 percent of the total area of the sphere.