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The distance between the slit and the biprism and between the biprims and the screen 50 cm each. The angle of the biprism is `179^(@)` and its refractive index is `1.5`. If the distance between successive bright fringes is `0.0135 cm`, then the wavelenghts of lights isA. `5893xx10^(-10) cm`B. `5898xx10^(-8) cm`C. `5898xx10^(-8) m `D. `2946xx10^(-8) cm` |
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Answer» Correct Answer - B Here, `alpha+179+ alpha=180^(@)` `:. Alpha=0.5^(@)=(0.5xx pi)/(180) "rad"` Thus, `d= 2u (mu-1)alpha` `=2xx0.5(1.5-1)(0.5 xxpi)/(180)` `=0.004361m` `:. Lambda=(d)/(D) beta=(0.004361xx0.00135xx10^(-2))/(1)` `=5893Å=5893xx10^(-8) cm` |
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