Optics 4 Question 3

3. In an experiment for determination of refractive index of glass of a prism by iδ plot, it was found that a ray incident at an angle 35 suffers a deviation of 40 and that it emerges at an angle 79. In that case, which of the following is closest to the maximum possible value of the refractive index?

(2016 Main)

(a) 1.5

(b) 1.6

(c) 1.7

(d) 1.8

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Answer:

Correct Answer: 3. (a)

Solution:

δ=(i1+i2)A

40=(35+79)A

A=74

Now, we know that

μ=sin(A+δm2)sin(A2)

It we take the given deviation as the minimum deviation then,

μ=sin(74+402)sin(742)=1.51

The given deviation may or may not be the minimum deviation. Rather it will be less than this value. Therefore, μ will be less than 1.51.

Hence, maximum possible value of refractive index is 1.51 .



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