JEE MAIN - Physics (2022 - 28th July Morning Shift - No. 14)
As shown in the figure, after passing through the medium 1 . The speed of light $$v_{2}$$ in medium 2 will be :
$$\left(\right.$$ Given $$\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$$ )
_28th_July_Morning_Shift_en_14_1.png)
$$1.0 \times 10^{8} \mathrm{~ms}^{-1}$$
$$0.5 \times 10^{8} \mathrm{~ms}^{-1}$$
$$1.5 \times 10^{8} \mathrm{~ms}^{-1}$$
$$3.0 \times 10^{8} \mathrm{~ms}^{-1}$$
Explanation
$$V = {1 \over {\sqrt {\mu \varepsilon } }} = {1 \over {\sqrt {{\mu _r}{\varepsilon _r}{\mu _0}{\varepsilon _0}} }}$$
$$ \Rightarrow {V_2} = {c \over {\sqrt 9 }} = {10^8}$$ m/s
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