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If the earth is moving towards a stationary star at a speed of 30 kilometres per second, find the apparent wavelength oflight emitted from thestar. Therealwavelength hasthevalue 5875 A. |
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Answer» Solution :Here earth (observer) is approaching the star and hence, the mutual distance is decreasing. So, the observer will notice an increase in frequency or decrease in WAVELENGTH. If V be the relative velocity ofthe SOURCE and c, the velocity of light, then the change in wavelength is GIVEN by `Deltalambda = (v)/(c) xx lambda` Substituting the given values, we have `Deltalambda = (30xx10^(3))/(3 xx10^(8)) xx 5875 A` =` 5875 xx 10^(-4) A` Altered wavelength `lambda' = lambda - Deltalambda` ` = 5875 A - (5875 xx 10^(-4) A)` `lambda' = 0.9999 xx 5875 A` = 5874.4125 A |
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