A sheet of cellophane acts as a half-wave plate for light of wavelength 4000Å. If the index of refraction did not change with wavelength, how would the sheet behave with respect to light of wavelength 8000Å?
Correct Answer - 1.45
The path difference is `2mu t`.
Now for destructive interface it can be
`2mut=(lambda)/(2)` or `(3lambda)/(2)` or `(5lambda)/(2)` and so on………
`mu=(lambda)/(4 t),(3lambda)/(4 t),(5lambda)/(4 t)……..`
`=(580xx10^(-9))/(4xx0.3xx10^(-6)) ,...
Correct Answer - (1) `E=sigma/(2in_(0))` perpendiculary away from the sheet
`B=(mu_(0)sigmav)/2` parallel to the sheet in `-ve` and `+ve` y direction
(2) E= same as in (1);`B=0`
Correct Answer - decrease in energy is `(1)/(2)(epsilon_(0)AV^(2))/(d)((epsilon_(r)-1)/(epsilon_(r)));` energy is transferred to the agent introducing the slab