A) the wavelength and frequency both remain unchanged
B) the wavelength is doubled and the frequency remains unchanged
C) the wavelength is doubled and the frequency becomes half
D) the wavelength is halved and the frequency remains unchanged.
Correct Answer: D
Solution :
[d]: Frequency remains unchanged with change of medium. Refractive index, \[n=\frac{c}{v}=\frac{1/\sqrt{{{\varepsilon }_{0}}{{\mu }_{0}}}}{1/\sqrt{\varepsilon \mu }}=\sqrt{{{\varepsilon }_{r}}{{\mu }_{r}}}\] Since \[{{\mu }_{r}}\] is very close to \[1,n=\sqrt{{{\varepsilon }_{r}}}=\sqrt{4}=2\] Thus, tedium \[{{\lambda }_{medium}}=\frac{\lambda }{n}=\frac{\lambda }{2}\]
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