A) wavelength is halved and frequency remains unchanged
B) wavelength is doubled and frequency becomes half
C) wavelength is doubled and the frequency remains unchanged
D) wavelength and frequency both remain unchanged.
Correct Answer: A
Solution :
[a] Frequency remains constant during refraction \[{{v}_{med}}=\frac{1}{\sqrt{{{\mu }_{0}}{{\in }_{0}}\times 4}}=\frac{c}{2}\] \[\frac{{{\lambda }_{med}}}{{{\lambda }_{air}}}=\frac{{{v}_{med}}}{{{v}_{air}}}=\frac{c/2}{c}=\frac{1}{2}\] \[\therefore \]wavelength is halved and frequency remains unchanged.
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