JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-I

  • question_answer
    A monochromatic beam of light has a frequency \[v=\frac{3}{2\pi }\times {{10}^{12}}Hz\] and is propagating along the direction\[\frac{\widehat{i}+\widehat{j}}{\sqrt{2}}\]. It is polarized along \[\widehat{k}\] the direction. The acceptable form for the magnetic field is:   [JEE Online 15-04-2018]

    A) \[k\frac{{{E}_{0}}}{C}\left( \frac{\widehat{i}-\widehat{j}}{\sqrt{2}} \right)\cos \left[ {{10}^{4}}\left( \frac{\widehat{i}-\widehat{j}}{\sqrt{2}} \right).\overrightarrow{r}-(3\times {{10}^{12}})t \right]\]

    B) \[\frac{{{E}_{0}}}{C}\left( \frac{\widehat{i}-\widehat{j}}{\sqrt{2}} \right)\cos \left[ {{10}^{4}}\left( \frac{\widehat{i}+\widehat{j}}{\sqrt{2}} \right).\overrightarrow{r}-(3\times {{10}^{12}})t \right]\]

    C) \[\frac{{{E}_{0}}}{C}\widehat{k}\cos \left[ {{10}^{4}}\left( \frac{\widehat{i}+\widehat{j}}{\sqrt{2}} \right).\overrightarrow{r}+(3\times {{10}^{12}})t \right]\]

    D) \[\frac{{{E}_{0}}}{C}\frac{\left( \widehat{i}+\widehat{j}+\widehat{k} \right)}{\sqrt{3}}\cos \left[ {{10}^{4}}\left( \frac{\widehat{i}+\widehat{j}}{\sqrt{2}} \right).\overrightarrow{r}+(3\times {{10}^{12}})t \right]\]

    Correct Answer: A

    Solution :

    Direction of B is, \[=\widehat{K}\times \widehat{E}\] \[=(\frac{\widehat{i}+\widehat{j}}{\sqrt{2}})\times \widehat{K}\] \[=\frac{\widehat{i}+\widehat{k}}{\sqrt{2}}+\frac{\widehat{j}\times \widehat{k}}{\sqrt{2}}\] \[=\frac{\widehat{j}}{\sqrt{2}}+\frac{\widehat{i}}{\sqrt{2}}=\frac{\widehat{i}-\widehat{j}}{\sqrt{2}}\] So, ans is between (i) and (ii) Propagation direction, \[\widehat{k}=\frac{\widehat{i}+\widehat{j}}{\sqrt{2}}\] \[\overrightarrow{B}\] wave will be \[\Rightarrow \frac{{{E}_{0}}}{c}(\widehat{B})Cos[|k|\widehat{k}-wt]\] \[\begin{matrix}    \downarrow  & \downarrow   \\    \text{we know it is} & \text{we know it is}  \\ \end{matrix}\] \[\frac{\widehat{i}-\widehat{j}}{\sqrt{2}}\]                          \[\frac{\widehat{i}+\widehat{j}}{\sqrt{2}}\] Only (i) satisfies Hence, correct answer is (i)


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