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CBSE Class 12 Physics 2017 Delhi Set 1 Paper

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Question : 20 of 26
Marks: +1, -0
(i) A ray of light incident on face ABA B of an equilateral glass prism, shows minimum deviation of 3030^{\circ}. Calculate the speed of light through the prism.
(ii) Find the angle of incidence at face ABA B so that the emergent ray grazes along the face AC.
Solution:  
(i) Calculation of speed of light
(ii) Calculation of angle of incidence at face AB
(i) μ=sin  (A+δm2)sin  (A2)\mu = \frac{ \sin\; \left( \frac{A+\delta_m}{2} \right) }{ \sin\; \left( \frac{A}{2} \right) }
  =  sin  (60+302)sin  (602)=2\;=\; \frac{ \sin\; \left( \frac{60^{\circ}+30^{\circ}}{2} \right) }{ \sin\; \left( \frac{60^{\circ}}{2} \right) } = \sqrt{2}
   Also       μ  =  cvv=3×1082m/s\;\text{ Also }\; \;\; \mu\;=\;\frac{c}{v} \Rightarrow v = \frac{3 \times 10^8}{\sqrt{2}} \text{m/s}
v  =2.122×108m/sv\;= 2.122 \times 10^8 \text{m/s}
(ii)
At face ACAC, let the angle of incidence be r2r_2.
For grazing ray, e=90e = 90^{\circ}
    μ=1sin  r2r2=sin1(12)=45\Rightarrow \;\; \mu = \frac{1}{\sin\; r_2} \Rightarrow r_2 = \sin^{-1}\left( \frac{1}{\sqrt{2}} \right) = 45^{\circ}
Let angle of refraction at face ABAB be r1r_1.
Now r1+r2=Ar_1 + r_2 = A
    r1=Ar2=6045=15\therefore \;\; r_1 = A - r_2 = 60^{\circ} - 45^{\circ} = 15^{\circ}
Let angle of incidence at this face be ii
  μ  =  sin  isin  r1\;\mu\;=\;\frac{ \sin\; i }{ \sin\; r_1 }
    μ=2=sin  isin  15\Rightarrow \;\; \mu = \sqrt{2} = \frac{ \sin\; i }{ \sin\; 15^{\circ} }
  i=sin1(2sin  15)\therefore \; i = \sin^{-1}\left( \sqrt{2} \cdot \sin\; 15^{\circ} \right)
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