Question
Question: In an experiment for determination of refractive index of glass of a prism by \(i - \delta \) plot, ...
In an experiment for determination of refractive index of glass of a prism by i−δ plot, it was found that a ray incident at angle 35∘ suffers a deviation of 40∘and that it emerges at angle 79∘. In that case, which of the following is closest to the maximum possible value of the refractive index?
A) 1.6
B) 1.7
C) 1.8
D) 1.5
Solution
Hint The first relation that we will be using in this question is : δ=i+e−A
Where: δ is the angle of deviation.
i is the angle of incidence.
e is the angle of emergence.
Ais the angle of prism.
We are given the values of δ, i and e. So, we get the value of A.
Value of refractive index in terms of A is
μmax=sin(2A)sin(2A+δ)
We have the values of A and δ, we can easily get the value of μ.
Complete step-by-step solution :
We are given
Angle of deviation, δ=40∘
Angle of incidence, i=35∘
Angle of emergence, e=79∘
Now, using the relation
δ=i+e−A
And putting in the values, we get
40∘=35∘+79∘−A A=74∘
So, angle of prism, A=74∘
Refractive index (μ) is given in terms of angle of prism (A) and angle of deviation (δ) as:
μmax=sin(2A)sin(2A+δ)
Putting in the values of A and δ, we get
μmax=sin(274∘)sin(274∘+40∘)=sin37∘sin57∘ =0.60180.8386=1.4
The actual formula for μmaxinvolves the value of δmin and not δ. We have assumed that δmin=40∘ which may not be true. Hence, maximum possible value of μ will be closest to 1.5
Note:- Since, we are to find the maximum possible values of μ, we have used approximations such as substituting δmin=40∘. One should take care during calculations especially, while dealing with sine values.