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Refractive index of prism of prism angle 6^@ is 1.5, then minimum deviation is ...... .

Answer»

`3^@`
`6^@`
`2^@`
`1^@`

Solution :Refractive index, n = `(SIN((A+delta_m)/(2)))/(sin(A/2))`
For SMALL values of A and `delta_m`
`sin((A+delta_m)/(2))`
`=(A+delta_m)/(2)` and `sin(A/2)~~A/2`
`n=((A+delta_m)/(2))/(A/2)`
`THEREFORE n=(A+delta_m)/(A)impliesnA-A=delta_m`
`therefore delta_m=(n-1)A therefore delta_m=(1.5-1)6^@therefore delta_m=3^@`


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