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CBSE Class 12 Physics 2023 Outside Delhi Set 3 Paper

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Question : 11 of 13
Marks: +1, -0
SECTION - C
A ray of light is refracted by a glass prism. Obtain an expression for the refractive index of the glass in terms of the angle of prism AA and the angle of minimum deviation 8 m8\,\text{m}.
Solution:     👈: Video Solution
A ray is refracted by a glass prism.
Angle of deviation =δ=i1+i2+A=\delta=i_1+i_2+A
Angle of prism =A=r1+r2=A=r_1+r_2
For minimum deviation,
  i1  =i2   and   r1=r2\; i_1\; = i_2 \;\text{ and }\; r_1=r_2
  ∴  δmin⁡  =2i1−A\; \therefore \; \delta_{\min} \; = 2 i_1 - A
  ∴  i1  =  A+δm2\; \therefore \; i_1\; = \; \frac{A+\delta_m}{2}
     Also,     A  =2r1\; \; \text{ Also, }\; \; A\; = 2 r_1
  ∴  r1  =  A2\; \therefore \; r_1\; = \; \frac{A}{2}
  ∴     Refractive index     =μ\; \therefore \; \; \text{ Refractive index }\; \; = \mu
  =  sin⁡i1sin⁡r1=  sin⁡A+δm2sin⁡A2\; = \; \frac{ \sin i_1 }{ \sin r_1 } = \; \frac{ \sin \frac{A+\delta_m}{2} }{ \sin \frac{A}{2} }
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