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=35^(@)`, `e=79^(@)`, `delta=40^(@)`
we know `delta=i+e-AimpliesA=i+e-delta`
`A=35^(@)+79^(@)-40^(@)=74^(@)`
Refractive index of prism `mu=(sin((A+delta_(m))/(2)))/(sin((A)/(2)))`
`mu=(sin(37^(@)+(delta_(m))/(2)))/(sin37^(@))=(5)/(3)sin(37^(@)+(delta_(m))/(2))`
Maximum value of `mu` can be `(5)/(3)`, so required
value of `mu` should be less than `(5)/(3)`
also `delta_(m)` will be less than `40^(@)`, so
`mult(5)/(3)sin(37^(@)+(40^(@))/(2))=(5)/(2)sin57^(@)`
`mult(5)/(3)sin57^(@)lt(5)/(3)sin60^(@)=1.44`
`mult1.44`
so the nearest possible value of `mu` for the given arrangement should be `1.5`