In a spectrometer experiment, the angle of minimum deviation was found to be `48.6^@`. What is the percentage accuracy in the measurement of refractive index of the prism ? Given least count of spectrometer `= 0.2^@` and angle of prism `= 60^@`.

Explanation

Here, `A = 60^(@), delta_(m) = 48.6^(@)`,
`mu = (sin(A + delta_(m))//2)/(sin A//2) = 1.6242`
As least count of spectrometer `= 0.2^(@)`,
Therefore, error in measurement of angle `A = +- 0.2^(@)`
Taking only + sign
`A' = (60 + 0.2) = 60.2^(@)` ,
`delta_(m)' = 48.6 + 0.2 = 48.8^(@)`
`:. mu +- dmu = (sin(A' + delta_(m)')//2)/(sin A'//2) = 1.6233`
`d mu = 1.6242 - 1.6233 = 0.0009`
`%` age accuracy
`= (d mu)/(mu) xx 100 = (0.0009)/(1.6242) xx 100 = 0.0554%`
Try yourself, taking negative sign of the error in measuring A and `delta_(m)`.