Word Count : 417

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बहुत बढ़िया — चलिए इसे हाथ से (manually) कदम-दर-कदम हल करते हैं। दिया हुआ विद्युत् क्षेत्र: $$ \vec E=(800\ \text{V/m})\,\hat x\;\sin\big(2\pi\times 10^{8}\,t + k z\big). $$ ### 1) आवृत्ति $f$ आर्गुमेंट में $ \omega t = 2\pi\times 10^8\,t$ है, अतः $$ \omega = 2\pi\times 10^{8}\ \text{rad/s}. $$ फ्रिक्वेंसी (f=\dfrac{\omega}{2\pi}=\dfrac{2\pi\times10^{8}}{2\pi}=10^{8}\ \text{Hz}=100\ \text{MHz}. --- ### 2) तरंगदैर्ध्य $\lambda$ और तरंगसंख्या $k$ खाली आकाश (vacuum) में तरंग की चाल $c=3.0\times10^{8}\ \text{m/s}$ ली जाए। $$ \lambda=\dfrac{c}{f}=\dfrac{3.0\times10^{8}}{1.0\times10^{8}}=3.0\ \text{m}. $$ तरंग संख्या (मापदंड) का परिभाषा $\displaystyle k=\dfrac{2\pi}{\lambda}$ या $\ ______ ______ ___ _________ ___ ___ ________ _____ ___ _________ _________ __________.
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