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 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?

If μ is refractive index of material of prism, then from Snell's lawμ=sinrsini=sinA/2sin(A+δm)/2where, A is angle of prism and δm is minimum deviation through prism.Given, i=35∘,δ=40∘,e=79∘.So, angle of deviation by a glass prism,δ=i+e−A⇒40∘=35∘+79∘−Ai.e. Angle of prism ⇒A=74∘.Such that, r1+r2=A=74∘.Let us put μ=1.5 in eq. (i), we get1.5=sinA/2sin(2A+δmin)⇒1.5=sin37∘sin(274∘+δmin)⇒0.9=sin(37∘+2δmin)(∵sin37∘≈0.6)sin64∘=sin(37∘+2δmin)(∵sin64∘=0.9)37∘+2δmin=64∘⇒δmin≈54∘This angle is greater than the 40∘ deviation angle already given. For greater μ, deviation will be even higher. Hence, μ of the given prism should be lesser than 1.5. Hence, the closest option will be 1.5.

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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 an angle $3 5^{\circ}$ suffers a deviation of $4 0^{\circ}$ and that it emerges at an angle $7 9^{\circ}$ . In that case, which of the following is closest to the maximum possible value of the refractive index?

1.5

1.6

1.7

1.8

Views: 6,072 students

Updated on: May 14, 2023

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2016

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Alexander Discussed

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 an angle

3 5^{\circ}

suffers a deviation of

4 0^{\circ}

and that it emerges at an angle

7 9^{\circ}

. In that case, which of the following is closest to the maximum possible value of the refractive index?

6 mins ago

Discuss this question LIVE

6 mins ago

Text solutionVerified

μ

is refractive index of material of prism, then from Snell's law

μ = \frac{s i n i}{s i n r} = \frac{s i n ( A + δ _{m} ) /2}{s i n A /2}

where,

A

is angle of prism and

δ_{m}

is minimum deviation through prism.
Given,

i = 3 5^{\circ}, δ = 4 0^{\circ}, e = 7 9^{\circ}

.
So, angle of deviation by a glass prism,

δ = i + e - A \Rightarrow 4 0^{\circ} = 3 5^{\circ} + 7 9^{\circ} - A

i.e. Angle of prism

\Rightarrow A = 7 4^{\circ}

.
Such that,

r_{1} + r_{2} = A = 7 4^{\circ}

.
Let us put

μ = 1.5

in eq. (i), we get

1.5 = \frac{s i n ( \frac{A + δ _{m i n}}{2} )}{s i n A /2}

\Rightarrow 1.5 = \frac{s i n ( \frac{7 4 ^{\circ} + δ _{m i n}}{2} )}{s i n 3 7 ^{\circ}}

\Rightarrow 0.9 = sin (3 7^{\circ} + \frac{δ _{m i n}}{2}) (∵ sin 3 7^{\circ} \approx 0.6)

sin 6 4^{\circ} = sin (3 7^{\circ} + \frac{δ _{m i n}}{2}) (∵ sin 6 4^{\circ} = 0.9)

3 7^{\circ} + \frac{δ _{m i n}}{2} = 6 4^{\circ} \Rightarrow δ_{m i n} \approx 5 4^{\circ}

This angle is greater than the

4 0^{\circ}

deviation angle already given. For greater

μ

, deviation will be even higher. Hence,

μ

of the given prism should be lesser than 1.5. Hence, the closest option will be 1.5.

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Question Text	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 an angle $3 5^{\circ}$ suffers a deviation of $4 0^{\circ}$ and that it emerges at an angle $7 9^{\circ}$ . In that case, which of the following is closest to the maximum possible value of the refractive index?
Updated On	May 14, 2023
Topic	Ray Optics and Optical Instruments
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 3
Upvotes	270
Avg. Video Duration	7 min

Text solutionVerified

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