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 +- d mu = (sin(A' + delta_m')//2)/(sin A'//2) = 1.6233`
`d mu = 1.6342 - 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`.