JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 1)
An electromagnetic wave of intensity 50 Wm–2 enters in a medium of refractive index 'n' without any loss. The ratio of the magnitudes of electric, and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by:
$$\left( {{1 \over {\sqrt n }},{1 \over {\sqrt n }}} \right)$$
$$\left( {\sqrt n ,\sqrt n } \right)$$
$$\left( {\sqrt n ,{1 \over {\sqrt n }}} \right)$$
$$\left( {{1 \over {\sqrt n }},\sqrt n } \right)$$
คำอธิบาย
C $$ = {1 \over {\sqrt {{\mu _0}{ \in _0}} }}$$
V = $$ = {1 \over {\sqrt {k{ \in _0}{\mu _0}} }}$$ [For transparent medium $$\mu $$r $$ \approx $$ $$\mu $$0]
$$ \therefore $$ $${C \over V}$$ $$=$$ $$\sqrt k = $$ n
$${1 \over 2} \in {}_0\,E_0^2$$C $$=$$ intensity $$=$$ $${1 \over 2}$$$$ \in $$0 kE2v
$$ \therefore $$ E$$_0^2$$C $$=$$ kE2v
$$ \Rightarrow $$ $${{E_0^2} \over {{E^2}}} = {{kV} \over C} = {{{n^2}} \over n} \Rightarrow {{{E_0}} \over E} = \sqrt n $$
similarly
$${{B_0^2C} \over {2{\mu _0}}} = {{{B^2}v} \over {2{\mu _0}}} \Rightarrow {{{B_0}} \over B} = {1 \over {\sqrt n }}$$
V = $$ = {1 \over {\sqrt {k{ \in _0}{\mu _0}} }}$$ [For transparent medium $$\mu $$r $$ \approx $$ $$\mu $$0]
$$ \therefore $$ $${C \over V}$$ $$=$$ $$\sqrt k = $$ n
$${1 \over 2} \in {}_0\,E_0^2$$C $$=$$ intensity $$=$$ $${1 \over 2}$$$$ \in $$0 kE2v
$$ \therefore $$ E$$_0^2$$C $$=$$ kE2v
$$ \Rightarrow $$ $${{E_0^2} \over {{E^2}}} = {{kV} \over C} = {{{n^2}} \over n} \Rightarrow {{{E_0}} \over E} = \sqrt n $$
similarly
$${{B_0^2C} \over {2{\mu _0}}} = {{{B^2}v} \over {2{\mu _0}}} \Rightarrow {{{B_0}} \over B} = {1 \over {\sqrt n }}$$