JEE MAIN - Physics (2018 - 15th April Evening Slot - No. 2)
A plane polarized monochromatic EM wave is traveling in vacuum along z direction such that at t = t1 it is found that the electric field is zero at a spatial point z1. The next zero that occurs in its neighbourhood is at z2. The frequency of the electroagnetic wave is :
$${{3 \times {{10}^8}} \over {\left| {{z_2} - {z_1}} \right|}}$$
$${{1.5 \times {{10}^8}} \over {\left| {{z_2} - {z_1}} \right|}}$$
$${{6 \times {{10}^8}} \over {\left| {{z_2} - {z_1}} \right|}}$$
$${1 \over {{t_1} + {{\left| {{z_2} - {z_1}} \right|} \over {3 \times {{10}^8}}}}}$$
Explanation
Since $$c = f\lambda \Rightarrow f = {c \over \lambda }$$. Here, $$\lambda = 2|{z_2} - {z_1}|$$ and $$c = 3 \times {10^8}$$. Therefore,
$$f = {{3 \times {{10}^8}} \over {2|{z_2} - {z_1}|}} = {{1.5 \times {{10}^8}} \over {|{z_2} - {z_1}|}}$$
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