JEE MAIN - Physics (2021 - 24th February Evening Shift - No. 21)
An electromagnetic wave of frequency 3 GHz enters a dielectric medium of relative electric permittivity 2.25 from vacuum. The wavelength of this wave in that medium will be _________ $$\times$$ 10$$-$$2 cm.
Answer
667
Explanation
Given, frequency of wave, f = 3 GHz = 3 $$\times$$ 109 Hz
Relative permittivity, $$\varepsilon $$r = 2.25
Since, f = C/$$\lambda$$
$$ \Rightarrow \lambda = {c \over f} = {{3 \times {{10}^8}} \over {3 \times {{10}^9}}} = 0.1$$ m
$$\because$$ $$\lambda$$m (wavelength of wave in a medium) = $$\lambda$$/$$\mu$$ and we know that, $$\mu = \sqrt {{\mu _r}{\varepsilon _r}} $$
As, dielectric is non-magnetic, $$\mu$$r = 1
$$ \Rightarrow \mu = \sqrt {2.25} = 1.5$$
$$ \Rightarrow {\lambda _m} = {{0.1} \over {1.5}} = {1 \over {15}} = 0.0667$$ m
= 6.67 cm = 667 $$\times$$ 10$$-$$2 cm
Relative permittivity, $$\varepsilon $$r = 2.25
Since, f = C/$$\lambda$$
$$ \Rightarrow \lambda = {c \over f} = {{3 \times {{10}^8}} \over {3 \times {{10}^9}}} = 0.1$$ m
$$\because$$ $$\lambda$$m (wavelength of wave in a medium) = $$\lambda$$/$$\mu$$ and we know that, $$\mu = \sqrt {{\mu _r}{\varepsilon _r}} $$
As, dielectric is non-magnetic, $$\mu$$r = 1
$$ \Rightarrow \mu = \sqrt {2.25} = 1.5$$
$$ \Rightarrow {\lambda _m} = {{0.1} \over {1.5}} = {1 \over {15}} = 0.0667$$ m
= 6.67 cm = 667 $$\times$$ 10$$-$$2 cm
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