A ray OP of monochromatic light is incident on the face AB of prism ABCD mear vertex B at an incident angle of `60degree` (see figure). If the refractive index of the material of the prism is `sqrt3`, which of the following is (are) are correct? `

Options

A. The ray gets totally internally reflected at face CD
B. The ray comes out through face AD
C. The angle between the incident ray and the emergent ray if `90degree`
D. The angle between the incident ray and the emergent ray is `120degree`

Answer

A, B, C, D

Explanation

Applying Snell's law at P
`n_(1)sin60degree=n_(2)sintheta`
`implies` `sin60degree=sqrt3sintheta` `implies` `theta=30degree`
In quadrilateral BCQP,
`60degree+(90degree+30degree)+135degree+anglePQC=360degree`
`implies` `anglePQC=45degree` `implies` `i=45degree`
The critical angle for prism - air pair of media is `C=sin^(-1)((1)/(sqrt3))` which is less than `45degree`.
Therefore total internal reflection takes place at face CD.
option (a) is correct.
In `triangle`QDM`,`angleQMD+180degree-(45degree+75degree)=60degree`
Therefore the angle of incidence of ray QM on AD is `30degree`.
This angle is less than the critical angle. Therefore the ray emerges out of face AD.
Option (b) is correct.
Applying Snell's law at M, we get
`sqrt3sin30degree=1xxsinalphaimpliesalpha=sin^(-1)((sqrt3)/(2))=60degree`
In quadrilateral PQMN, `anglePNM=360degree-[60degree+90degree+120degree]=90degree`
`:'` The angle between the incident ray and the emergent ray is `90degree` Option (c) is correct.