Consider a light ray travelling in air is incident into a medium of refractive index √2 n. The incident angle is twice that of refracting angle. Then, the angle of incidence will be :

Question

Consider a light ray travelling in air is incident into a medium of refractive index √2 n. The incident angle is twice that of refracting angle. Then, the angle of incidence will be : (A)  sin -1(√n) (B)  cos -1(√(n/2)) (C)  sin -1(√2 n) (D) 2 cos -1(√(n/2))

Tardigrade Admin · Accepted Answer

i=2 r sin i × n1= sin r × n2 sin i × 1= sin (i/2) × √2 n ( sin i/ sin (i)2)=√2 n (2 sin (i/2) cos (i/2)/ sin (i)2)=√2 n cos (i/2)=√(n/2) (i/2)= cos -1(√(n/2)) i=2 cos -1(√(n/2))

Q. Consider a light ray travelling in air is incident into a medium of refractive index $2 n$ . The incident angle is twice that of refracting angle. Then, the angle of incidence will be :

$sin^{- 1} (n)$

$cos^{- 1} (\frac{n}{2})$

$sin^{- 1} (2 n)$

$2 cos^{- 1} (\frac{n}{2})$

Q. Consider a light ray travelling in air is incident into a medium of refractive index 2n​. The incident angle is twice that of refracting angle. Then, the angle of incidence will be :

sin−1(n​)

cos−1(2n​​)

sin−1(2n​)

2cos−1(2n​​)

i=2r sini×n1​=sinr×n2​ sini×1=sin2i​×2n​ sin2i​sini​=2n​ sin2i​2sin2i​cos2i​​=2n​ cos2i​=2n​​ 2i​=cos−1(2n​​) i=2cos−1(2n​​)

Q. Consider a light ray travelling in air is incident into a medium of refractive index $2 n$ . The incident angle is twice that of refracting angle. Then, the angle of incidence will be :

$sin^{- 1} (n)$

$cos^{- 1} (\frac{n}{2})$

$sin^{- 1} (2 n)$

$2 cos^{- 1} (\frac{n}{2})$