1.

Minimum deviation of prism having refractive index mu and small angle of prism A is shown by .....

Answer»

`delta_m=(mu-1)A`
`delta_m=A(mu+1)`
`delta=(SIN((A+delta_m)/(2)))/(sintheta/2)`
`delta_m=A[(mu-1)/(mu+1)]`

SOLUTION :`mu=(sin((A+delta_m)/(2)))/(sin(theta)/(2))`
For small ANGLE of PRISM and `delta_m`
`sin((A+delta_m)/(2))~~(A+delta_m)/(2)`
and `sin(A/2)~~A/2`
`THEREFORE mu=(A+delta_m)/(2)`
`therefore muA=A+delta_m``therefore delta_m=(mu-1)A`


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