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Refraction Of Light Through A Prism

Question
ICSEENIPH12029779
Wired Faculty App

At what angle, a ray of light should be incident on the first face AB of a regular glass prism ABC so that the emergent ray grazes the adjacent face AC? See Figure 7 below. (Refractive Index of glass = 1.6)

Solution

straight mu presubscript straight a superscript straight g space equals space fraction numerator sin space straight i over denominator sin space straight r subscript straight i end fraction 1.6 space equals space fraction numerator sin space straight i over denominator sin space straight r subscript straight i end fraction straight mu presubscript straight g superscript straight a space equals space fraction numerator sin space straight r subscript 2 over denominator sin space 90 end fraction sin space straight r subscript 2 space equals space fraction numerator 1 over denominator 1.6 end fraction space space space space straight r subscript 2 space equals space 38.7 straight r subscript 1 space equals space 60 space minus space 38.7 space space space equals 21.3 1.6 space equals space fraction numerator sin space straight i over denominator sin space straight r end fraction space equals space 21.3 space space sin space straight i space equals space 0.5812 space straight i space equals space 35.5 space   

Question
ICSEENIPH12029827
Wired Faculty App

For any prism, show that refractive index of its material is given by:


straight n space or space straight mu space equals space fraction numerator sin space open parentheses begin display style fraction numerator straight A space plus straight delta subscript straight m over denominator 2 end fraction end style close parentheses over denominator sin space open parentheses begin display style straight A over 2 end style close parentheses end fraction 
where the terms have their usual meaning.

Solution


Here, 


i = Angle of incidence

e= Angle of emergence

A = refracting angle

r + r’ = Angle of a refraction

δ = deviation

i + e = A + δ                               ---(i)

A = r1+ r2                                              ---(ii)

Above relation connecting i. c., A + δ are shown above.

Under minimum deviation i = e, rt = r2 = r

From Eq. (i), we have

2 i = A + δ min

 i space equals space fraction numerator straight A space plus space straight delta subscript min over denominator 2 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis i v right parenthesis F r o m space e q n. left parenthesis i i right parenthesis A space equals 2 r r space equals A over 2 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis v right parenthesis mu space equals space fraction numerator s i n space i over denominator s i n space r end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis v i right parenthesis F r o m space E q. space left parenthesis i v right parenthesis comma space left parenthesis v right parenthesis space a n d space left parenthesis v i right parenthesis space w e space g e t mu space equals space fraction numerator S i n space open parentheses begin display style fraction numerator A space plus space delta subscript m i n end subscript over denominator 2 end fraction end style close parentheses over denominator s i n space open parentheses begin display style A over 2 end style close parentheses end fraction 
Hence proved. 

Question
ICSEENIPH12029853
Wired Faculty App

A ray of light LM incident normally on the surface AC of an isosceles right angled prism ABC (where AB = BC) emerges along PQ, parallel to LM, shwon in Figure.


What can you say about refractive index μ of the material of the prism?

Solution

∠A = angleC = 45°

∠ i = 45°

For total internal reflection to take place at faces,
sin space i space equals space 1 over n semicolon space w h e r e space n space i s space t h e space r e f r a c t i v e space i n d e x space o f space p r i s m space w. r. t o space a i r s i n space 45 greater than thin space 1 over straight n space semicolon straight n space greater than thin space fraction numerator 1 over denominator sin space 45 end fraction straight n space greater than thin space square root of 2 Refractive space index space of space prism space is space greater space than space square root of 2

Question
ICSEENIPH12029873
Wired Faculty App

Calculate angle of minimum deviation (δm) for a regular glass prism. (Refractive index of glass = 1-6) 

Solution

Given,
n = 1.6

sin = ? 
We  know, 
n = fraction numerator sin space open parentheses begin display style fraction numerator straight A space plus space sin over denominator 2 end fraction end style close parentheses over denominator sin space straight A divided by 2 end fraction 
1.6 = fraction numerator sin space open parentheses begin display style fraction numerator 60 space plus space sin over denominator 2 end fraction end style close parentheses over denominator sin space open parentheses begin display style 60 over 2 end style close parentheses end fraction 
53.13 = fraction numerator 60 space plus space sin over denominator 2 end fraction 
   sin = 46.26o 

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