JEE MAIN - Physics (2020 - 8th January Morning Slot - No. 20)
The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 3 and relative permeability $${4 \over 3}$$ for this wavelength, will be :
45°
15°
30°
60°
Explanation
n = $$\sqrt {{\varepsilon _r}{\mu _r}} = \sqrt {3 \times {4 \over 3}} $$ = 2
n sin c = 1 sin 90o
sin c = $${1 \over 2}$$
$$ \Rightarrow $$ c = 30o
n sin c = 1 sin 90o
sin c = $${1 \over 2}$$
$$ \Rightarrow $$ c = 30o
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