b
\(\,\,{\lambda _1} = 3 \times {10^{ - 3}}m,\,\,{\lambda _2} = I\,m,\,c = 3 \times {10^8}m{s^{ - 1}}\)

\(\therefore \,\,{{f}_1} = \frac{c}{{{\lambda _1}}} = \frac{{3 \times {{10}^8}}}{{3 \times {{10}^{ - 3}}}} = {10^{11}}Hz = {10^5}MHz\,\,\,\& \,\,\)

\({{f}_2} = \frac{c}{{{\lambda _2}}} = \frac{{3 \times {{10}^8}}}{1} = 3 \times {10^8}Hz = 3 \times {10^2}MHz\)

તેથી અનુરૂપ આવૃતિ ગાળો  \(3 × 10^2\, MHz \) થી \(10^5 \,MHz\)