Question

According to the quantum theory, a photon of electromagnetic radiation of frequency $v$ has energy $E=h v$, 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 $m=\frac{h v}{c^{2}}$.
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. $\frac{g d^{2}}{2 c^{2}}$
• B. $\frac{h}{m c}$
• C. $\frac{m c d^{2}}{h}$
• 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,
$h=\frac{1}{2} g t^{2}=\frac{1}{2} \times g \times\left(\frac{d}{c}\right)^{2}$
$=\frac{g d^{2}}{2 c^{2}}$