Question

According to the quantum theory, a photon of electromagnetic radiation of frequency $v$ has energy $E=hv$, where $h$ is known as Planck's constant.
According to the theory of relativity, a particle of mass $m$ has equivalent energy $E=m{c}^2$, where $c$ is speed of light. Thus, a photon can be treated as a particle having effective mass $m=\frac{hv}{c^2}$.
If a flash of light is sent horizontally in earth's gravitational field, then photons while travelling a horizontal distance $d$ would fall through a distance given by

• A. $\frac{g{d}^2}{2c^2}$
• B. $\frac{h}{mc}$
• C. $\frac{mc{d}^2}{h}$
• D. zero

Answer 1

Answer: (A)

In time $t$, a particle of mass $m$ falls by a distance
$h=\frac{1}{2}g{t}^2$
Now, distance covered horizontally $=d$
and speed in horizontal direction $=c$.
Time to travel distance, $d=t=\frac{d}{c}$

So, photons falls through a distance,
$h=\frac{1}{2}g{t}^2=\frac{1}{2}\times g\times {\left(\frac{d}{c}\right)}^2$
$=\frac{g{d}^2}{2c^2}$