Electromagnetic Waves NEET प्रश्न हिंदी में — भौतिक विज्ञान (Physics) MCQ उत्तर सहित

दिए गए समीकरण की तुलना करने पर :

  $$\begin{gathered} {\mathrm{E}}_{\mathrm{y}}=2.5\frac{\mathrm{N}}{\mathrm{C}}\cos \left[\left(2\mathrm{\pi }\times {10}^6\frac{\mathrm{rad}}{\mathrm{s}}\right)\mathrm{t} -\left(\mathrm{\pi }\times {10}^{-2}\frac{\mathrm{rad}}{\mathrm{m}}\right)\mathrm{x}\right] \\ \mathrm{With} \mathrm{the} \mathrm{standard} \mathrm{equation} \\ {\mathrm{E}}_{\mathrm{y}}={\mathrm{E}}_{\mathrm{o}}\cos (\mathrm{\omega t}-\mathrm{kx}) \\ \mathrm{we} \mathrm{get} \\ \mathrm{\omega }=2\mathrm{\pi f}=2\mathrm{\pi }\times {10}^6 \\ \therefore \mathrm{f}={10}^6 \mathrm{Hz} \\ \mathrm{Moreover}, \mathrm{we} \mathrm{know} \mathrm{that} \\ \frac{2\mathrm{\pi }}{\mathrm{\lambda }}=\mathrm{k}=\mathrm{\pi }\times {10}^{-2} {\mathrm{m}}^{-1} \\ \Rightarrow \mathrm{\lambda }=200 \mathrm{m} \end{gathered}$$

अतः, तरंग धनात्मक x-दिशा के अनुदिश आवृत्ति ${10}^6 \mathrm{Hz}$ तथा 200 m की तरंगदैर्ध्य के साथ गतिमान है।

$$\begin{gathered} E=cB=3\times {10}^8\times \left(100\mu T\right)\sin \left[\right(2\times {10}^{15}{s}^{-1}\left)\right(t-x/c\left)\right] =3\times {10}^{10}\times \sin \left[\right(2\times {10}^{15}{s}^{-1}\left)\right(t-x/c\left)\right] \\ As the wave is propagating along the +x-direction, \overrightarrow{E}\times \overrightarrow{B} should be along +x-direction. \\ \overrightarrow{E}=\left(3\times {10}^{10}\mu V/m\right)\times \sin \left[\right(2\times {10}^{15}{s}^{-1}\left)\right(t-x/c\left)\right]\left(-\hat{k}\right) \end{gathered}$$

$\mathrm{Frequency} \mathrm{of} \mathrm{\gamma }-\mathrm{rays} \mathrm{is} \mathrm{more} \mathrm{than} \mathrm{both} \mathrm{the} \mathrm{frequencies} \mathrm{of} \mathrm{X}-\mathrm{rays} \mathrm{and} \mathrm{ultrav}i\mathrm{olet} \mathrm{rays}.$

$J=\frac{P}{A}=\frac{3.9\times {10}^{26}}{4\mathrm{\pi }\left(6.96\times {10}^8\right)}=6.4\times {10}^7 \frac{W}{m^2}$