Question

A glass ($\mu = 2$) prism is in the shape of a quarter cylinder of length $L$ and radius $R = 5$ cm, lying on a horizontal table. A uniform horizontal light beam fall on its vertical plane surfaces, as shown in figure. If a plane paper of width 5 cm and length $L$ is placed at a distance 7.5 cm from $O$, then the fraction of light (in percentage) falling on the prism that will fall come out of prism is ____.

• A. 12.5
• B. 25
• C. 30
• D. 50

Answer 1

Answer: (C)

30

Answer 2

The light beam falls on the vertical plane surface of the quarter cylinder. Since the rays are horizontal and the vertical surface is flat, the light enters the glass undeviated. The rays travel horizontally inside the glass and hit the curved surface. Let the origin O be at the bottom left corner. The vertical face is along the y-axis from y=0 to y=R. The curved surface is a quarter circle with equation $x^2 + y^2 = R^2$ in the first quadrant.

Consider a ray incident at height y on the vertical face. This ray travels horizontally and hits the curved surface at the point P with coordinates $(x, y)$ such that $x^2 + y^2 = R^2$. Since the ray is horizontal, its y-coordinate is y, and the x-coordinate is $x = \sqrt{R^2 - y^2}$.

At the curved surface, the light goes from glass ($\mu = 2$) to air ($\mu_{air} = 1$). The critical angle for total internal reflection (TIR) is $\theta_c$ such that $\sin \theta_c = \frac{\mu_{air}}{\mu} = \frac{1}{2}$. Thus, $\theta_c = 30^\circ$.

The normal to the curved surface at point P passes through the center O. The angle of incidence $i$ at the curved surface is the angle between the incident ray (horizontal) and the normal (radial). Let $\phi$ be the angle between the radius vector OP and the horizontal x-axis. Then $y = R \sin \phi$ and $x = R \cos \phi$. The normal at P makes an angle $\phi$ with the x-axis. Since the incident ray is horizontal, the angle of incidence is $i = \phi$.

For the light to come out of the prism, the angle of incidence must be less than the critical angle: $i < \theta_c$, which means $\phi < 30^\circ$.

Since $y = R \sin \phi$, the condition $\phi < 30^\circ$ is equivalent to $\sin \phi < \sin 30^\circ$, so $\frac{y}{R} < \frac{1}{2}$, which means $y < \frac{R}{2}$.

The incident light beam is uniform and falls on the vertical face of height R. The rays are incident at heights from y=0 to y=R. The rays incident at heights $0 \le y < R/2$ will come out of the prism. The rays incident at heights $R/2 \le y \le R$ will undergo TIR.

The fraction of light coming out of the prism is the ratio of the height range for which light comes out to the total height range of incidence:

Fraction = $\frac{R/2 - 0}{R - 0} = \frac{R/2}{R} = \frac{1}{2}$. In percentage, this fraction is $\frac{1}{2} \times 100\% = 50\%$.

A ray incident at height y, the point of exit is P$(\sqrt{R^2-y^2}, y)$. The normal at P makes an angle $\phi = \arcsin(y/R)$ with the x-axis. The angle of refraction is $r = \arcsin(2y/R)$.

The angle of the refracted ray with the x-axis is $\theta_{refracted}$. The normal is at angle $\phi$. The incident ray is horizontal. The angle of incidence is $i = \phi$. The angle of refraction is $r$. The refracted ray makes angle $r$ with the normal.

From the geometry, the angle between the refracted ray and the horizontal is $\alpha$. The normal makes an angle $\phi$ with the horizontal. The refracted ray makes an angle $r$ with the normal. So, $\alpha = r - \phi$. The ray goes downwards, so the angle with the positive x-axis is $-(\phi - r) = \phi - r$. The equation of the refracted ray is $Y - y = \tan(\phi - r) (X - x)$. The ray hits the x-axis when Y=0. So $-y = \tan(\phi - r) (X - x)$. $X = x - \frac{y}{\tan(\phi - r)}$. $x = R \cos \phi$, $y = R \sin \phi$. $X = R \cos \phi - \frac{R \sin \phi}{\tan(\phi - r)} = R \cos \phi - \frac{R \sin \phi \cos(\phi - r)}{\sin(\phi - r)}$. $X = R \frac{\cos \phi \sin(\phi - r) - \sin \phi \cos(\phi - r)}{\sin(\phi - r)} = -R \frac{\sin r}{\sin(\phi - r)}$. We have $\sin r = 2 \sin \phi$. $X = -R \frac{2 \sin \phi}{\sin(\phi - r)}$.

When $\phi \to 0$, $i \to 0$, $r \to 0$. $X \to R \cos 0 - \frac{R \sin 0}{\tan(0-0)} = R - 0/0$. Using L'Hopital's rule or series expansion, $\sin(\phi - r) \approx \phi - r$. $\sin r = 2 \sin \phi \approx 2 \phi$. So $r \approx 2 \phi$. $\phi - r \approx \phi - 2 \phi = -\phi$. $X \approx R \cos \phi - \frac{R \sin \phi}{\tan(-\phi)} = R \cos \phi + \frac{R \sin \phi}{\tan \phi} = R \cos \phi + R \cos \phi = 2 R \cos \phi$. As $\phi \to 0$, $X \to 2R$.

When $\phi \to 30^\circ$, $i \to 30^\circ$, $r \to \arcsin(2 \sin 30^\circ) = \arcsin(1) = 90^\circ$. $X = R \cos \phi - \frac{R \sin \phi}{\tan(\phi - r)}$. As $\phi \to 30^\circ$, $r \to 90^\circ$. $\phi - r \to 30^\circ - 90^\circ = -60^\circ$. $X \to R \cos 30^\circ - \frac{R \sin 30^\circ}{\tan(-60^\circ)} = R \frac{\sqrt{3}}{2} - \frac{R/2}{-\sqrt{3}} = R \frac{\sqrt{3}}{2} + \frac{R}{2\sqrt{3}} = R (\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{6}) = R (\frac{3\sqrt{3} + \sqrt{3}}{6}) = R \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3} R$. With R=5 cm, the range of x-coordinates where the light hits the table is from $2R = 10$ cm to $\frac{2\sqrt{3}}{3} R = \frac{10\sqrt{3}}{3} \approx \frac{10 \times 1.732}{3} \approx \frac{17.32}{3} \approx 5.77$ cm. So the light falls on the x-axis in the range $[5.77, 10]$.

The paper is placed from 7.5 cm to 12.5 cm. The overlap is from 7.5 to 10. The length is $10 - 7.5 = 2.5$. The total length where light comes out is $10 - \frac{10\sqrt{3}}{3} = 4.23$. The fraction of light coming out that falls on the paper is $\frac{2.5}{4.23} \approx 0.59$. The fraction of light falling on the prism that falls on the paper is $0.5 \times 0.59 = 0.295$. In percentage, 29.5%. Let's round to the nearest integer, which is 30.