Animated Solution for Physics - Optics: Screen $S$ is illuminated by two point sources $A$ and $B$. Another source $C$ sends a parallel beam of light towards point $P$ on the screen (see figure). Line $AP$ is normal to the screen and the lines $AP$, $BP$ and $CP$ are in one plane. The distances $AP$ and $BP$ are $3\text{ m}$ and $1.5\text{ m}$. The radiant powers of sources $A$ and $B$ are $90\text{ W}$ and $180\text{ W}$ respectively. The beam from $C$ is of intensity $20{\text{ W/m}}^2$. Calculate intensity at $P$ on the screen.

Have you ever wondered why a flashlight seems dimmer when you shine it at an angle against a wall compared to when you shine it straight on? Or why the sun feels hotter at noon than at sunset? The answer lies in the beautiful geometry of light, specifically a principle known as Lambert's Cosine Law. In this problem, we are going to explore exactly that by calculating the total intensity of light hitting a specific point on a screen from three different sources.

Imagine you are standing in a dark room, looking at a flat screen. Suddenly, three different light sources are turned on, all aiming at a single point $P$ on the screen.

Our first source, $A$, is a point source located $3\text{ m}$ directly in front of point $P$. Because it's directly in front, its light hits the screen perfectly head-on.

Our second source, $B$, is also a point source, but it's located $1.5\text{ m}$ away and strikes the screen at a $60^{\circ }$ angle from the normal (the perpendicular line to the screen).

Finally, we have source $C$. Unlike $A$ and $B$, source $C$ isn't a point source; it's a parallel beam of light, much like a laser pointer, with a given intensity of $20{\text{ W/m}}^2$. This beam also strikes the screen at a $60^{\circ }$ angle from the normal, but from the opposite side.

Because these three sources are independent, they are what physicists call incoherent. This is a crucial detail! It means their light waves don't interfere with each other to create complex bright and dark patterns. Instead, their energies simply add up. To find the total intensity at point $P$, we just need to calculate the individual intensity from each source and sum them together:

Before we calculate the individual intensities, we need to understand how light spreads out from a point source. Imagine a lightbulb emitting energy equally in all directions. As the light travels outward, it spreads over the surface of an expanding sphere. The surface area of a sphere is $4\pi r^2$, where $r$ is the distance from the source. Therefore, the intensity (power per unit area) at a distance $r$ is the total power $P$ divided by this area: $\frac{P}{4\pi r^2}$.

But there's a catch! This formula only works if the surface receiving the light is perfectly perpendicular to the light rays. If the surface is tilted, the same amount of light energy is spread over a larger area, making it less intense. This is where Lambert's Cosine Law comes in. We must multiply the intensity by the cosine of the angle of incidence $\theta$ (the angle between the light ray and the normal to the surface).

So, our master equation for the intensity from a point source on a flat screen is:

Now, let's apply our master equation to each source.

1. Intensity due to Source A ($I_A$) Source $A$ has a radiant power $P_A=90\text{ W}$ and is at a distance $r_A=3\text{ m}$. Because it lies on the normal line $AP$, the angle of incidence ${\theta }_A$ is $0^{\circ }$.

2. Intensity due to Source B ($I_B$) Source $B$ has a radiant power $P_B=180\text{ W}$ and is at a distance $r_B=1.5\text{ m}$. The light ray from $B$ makes an angle of $60^{\circ }$ with the normal $AP$, so ${\theta }_B=60^{\circ }$.

3. Intensity due to Source C ($I_C$) Source $C$ is a parallel beam, so we don't need to worry about the inverse-square law ($\frac{1}{r^2}$). Its intensity perpendicular to the beam is already given as $I_{C0}=20{\text{ W/m}}^2$. We only need to account for the tilt of the screen relative to the beam. The angle of incidence is ${\theta }_C=60^{\circ }$.

We've done the heavy lifting! We now know exactly how much intensity each source contributes to point $P$. The final step is the easiest: we simply add them together.

And there we have it! The total intensity at point $P$ on the screen is $13.97{\text{ W/m}}^2$. This problem is a fantastic reminder that complex physical situations can often be solved by breaking them down into their fundamental, independent components. By understanding how light spreads and how surfaces receive it, we can illuminate even the trickiest of physics problems!