Question

A laser is located at one focus of ellipse. A sheet of metal which is only a fraction of inch wide serves as reflecting surface,lines entire ellipse is located as same hight above the ground as laser. When beam stikes the metal surface i,t is reflected toword the other focus of ellipse. If focii are 30 ft apart. Shorter dimensions of ellipse is 14 feet . how much distance is travelled by beamfrom its point of occurance to the other focus.? Also caculate the time it will take to reach .(assume speed of laser =speed of light =$3\times {10}^8$

Answer 1

from above diagram, 2ae = 30, ae = 15, b=14

So,

$b^2=a^2-a^2{e}^2$

$a^2=b^2+a^2{e}^2$

$a^2=196+225$

$a^2=421$

$a=\sqrt{421}$

Ps = Ps'

Total distance covered by laser = 2Ps

$PS=\sqrt{{\left(OS\right)}^2+{\left(OP\right)}^2}$

$=\sqrt{{\left(15\right)}^2+{\left(14\right)}^2}$

$PS=\sqrt{421}$

Total distance covered by laser = $2\sqrt{421}$ ft.