There are \[{{n}_{1}}\] photons of frequency \[{{\gamma }_{1}}\] in a beam of light. In an equally energetic beam, there are \[{{n}_{2}}\] photons of frequency \[{{\gamma }_{2}}\]. Then the correct relation is [KCET 2003]

\[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{\gamma _{1}^{2}}{\gamma _{2}^{2}}\]

\[\frac{{{n}_{1}}}{{{n}_{2}}}=1\]

\[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{\gamma }_{1}}}{{{\gamma }_{2}}}\]

\[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{\gamma }_{2}}}{{{\gamma }_{1}}}\]

There are \[{{n}_{1}}\] photons of frequency \[{{\gamma }_{1}}\] in a

1.	There are \[{{n}_{1}}\] photons of frequency \[{{\gamma }_{1}}\] in a beam of light. In an equally energetic beam, there are \[{{n}_{2}}\] photons of frequency \[{{\gamma }_{2}}\]. Then the correct relation is [KCET 2003]
A.	\[\frac{{{n}_{1}}}{{{n}_{2}}}=1\]
B.	\[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{\gamma }_{1}}}{{{\gamma }_{2}}}\]
C.	\[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{\gamma }_{2}}}{{{\gamma }_{1}}}\]
D.	\[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{\gamma _{1}^{2}}{\gamma _{2}^{2}}\]
Answer» D. \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{\gamma _{1}^{2}}{\gamma _{2}^{2}}\]