Question

In an experiment for determination of refractive index of glass of a prism by `i-delta`, plot it was found thata ray incident at angle `35^circ`, suffers a deviation of `40^circ` and that it emerges at angle `79^circ`. In that case which of the following is closest to the maximum possible value of the refractive index?
A. `1.5`
B. `1.6`
C. `1.7`
D. `1.8`

Answer 1

Correct Answer - A
Here, `i_(1) = 35^(@), delta = 40^(@), i_(2) = 79^(@), mu = ?`
As `A + delta = i_(1) + i_(2)`
`A = (i_(1) + i_(2)) - delta = 35^(@) + 79^(@) - 40^(@) = 74^(@)`
`mu = (sin (A + delta_(m))//2)/(sin A//2) = (sin (74^(@) + 40^(@))//2)/(sin 74^(@)//2)`
`= (sin 57^(@))/(sin 37^(@)) = (0.8386)/(0.6018) = 1.4`
Here, we have assumed that `delta_(m) = 40^(@)` which may not be true. Hence maximum possible value of `mu` will be closest to `1.5`.