The angle of incidence for a ray of light at a refracting surface of a prism is `45^(@)`. The angle of prism is `60^(@)`. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are `:`

Options

A. `30^(@),sqrt2`
B. `45^(@),sqrt2`
C. `30^(@),(1)/(sqrt2)`
D. `45^(@),(1)/(sqrt2)`

Answer

A

Explanation

(a) Consider a ray of light PQ incident at the surface AB and moves along RS, after passing through the prism ABC.
It is given that the incident ray suffers minimum deviation. Therefore, the ray inside the prism must be parallel to the base BC of the prism.
From the geometry of the prism and the ray diagram, then it is clear that,
angle of incidence,i=`45^(@)`
angle of refraction, r=r'= `30^(@)`
angle of emergence, e= `45^(@)`
A= angle of prism=`60^(@)`
Therefore, minimum deviation suffered by the ray is
`delta_(min)=i+e-(r-r')=90^(@)-60^(@)=30^(@)`

Refractive index of the material,
`mu=sin((A+delta_(m))/2)/(sin""A/2)`
`therefore` `mu=sin((60^(@)+30^(@))/2)/(sin""(60^(@))/2)=(sin45^(@))/(sin30^(@))=(1//sqrt2)/(1//2)=2/(sqrt2)=sqrt2`