Solved Problem on Propagation of Light
advertisement   



A lamp of negligible dimensions is fixed on the ceiling of a room, whose height is 3 m. An opaque disc of 20 cm in diameter is suspended at 75 cm from the ceiling, their faces are horizontal and their center is in the same vertical as the lamp. Calculate the shadow area projected by the disc above the floor.


Problem data:
  • Disc diameter:    y = 20 cm;
  • Disk distance to the lamp:    p = 75 cm;
  • Distance from the shadow to the lamp:    q = 3 m.
Problem diagram:

Figure 1

Solution

First, we must convert the room height, given in meters (m), to centimeters (cm), rather than converting all other units to meters as usual.
\[ q=3\;\text{m}=3\times (100\;\text{cm})=300\;\text{cm} \]
To calculate the area of the shadow, first, we have to calculate the diameter of the shadow, this is given by the expression
\[ \bbox[#99CCFF,10px] {\frac{y'}{y}=\frac{q}{p}} \]
\[ \begin{gather} \frac{y'}{20}=\frac{300}{75}\\ y'=\frac{20\times 300}{75}\\ y'=\frac{6000}{75}\\ y'=80\;\text{cm} \end{gather} \]
The shadow radius is half the diameter.
\[ \begin{gather} r=\frac{y'}{2}\\ r=\frac{80}{2}\\ r=40\;\text{cm} \end{gather} \]
The area of a circle is given by
\[ \bbox[#99CCFF,10px] {A=\pi r^{2}} \]
assuming π = 3.14, the area of the shadow will be
\[ \begin{gather} A=3.14\times 40^{2}\\ A=3.14\times 1600 \end{gather} \]
\[ \bbox[#FFCCCC,10px] {A=5024\;\text{cm}^{2}} \]
advertisement   

Creative Commons License
Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .