A glass prism of angle `A = 60^(@)` gives minimum angle of deviation `theta ~~ 30^(@)` with the maximum error of `1^(@)` when a beam of parallel light passed through the prism during an experiment. Find the permissible error in the measurement of refractive index `mu` of the material of the prism.

Answer

`(5 pi)/(18) %`

Explanation

`theta=2i-2r=2i-A`
`rArri=(theta+A)/(2)=(60^(@)+30^(2))/(2)=45^(@)`
But `sin I = mu sin (A)/(2)`
`rArr sin((theta+A)/(2))=mu "sin"(A)/(2)rArr mu=sqrt(2)`
`mu=(sin((theta+A)/(2)))/("sin"(A)/(2))`
`rArr Delta mu=(cosec(A)/(2))cos((theta+A)/(2))((Deltatheta)/(2))B`
`=(cosec30^(@))cos45^(@)((1^(@))/(2))=(1)/(sqrt(2))((pi)/(180^(@)))`
`(Deltamu)/(mu)xx100={((pi)/(sqrt(2)(180)))(1)/(sqrt(2))}100=(5pi)/(18)%`