1.

The distance between the slit and the biprism and between the biprims and the screen50 cm each. The angle of the biprism is 179^(@) and itsrefractive index is 1.5. If the distance between successive bright fringes is 0.0135 cm, then the wavelenghts of lights is

Answer»

`5893xx10^(-10) cm`
`5898xx10^(-8) cm`
`5898xx10^(-8) m `
`2946xx10^(-8) cm`

Solution :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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