For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index

Options

A. lies between 2 and `sqrt2`
B. is less than 1
C. is greater than 2
D. lies between`sqrt `2 and 1

Answer

A

Explanation

`mu=(sin((delta_(m)+A)/(2)))/(sin((A)/(2)))`
`mu=(sin(A))/(sin((A)/(2)))=(2sin(A//2)cos(A//2))/(sin(A//2))`
`mu=2cos(A//2)`
`**i_("min")=0=A_("min")rArr mu_("min")=2`
`**i_("max")=pi//2=A_("max")rArr mu_("max")=sqrt2`
range of `mu` between `sqrt2 lt mu lt 2`
`delta_(m)=i+e-A`
`**"Given "delta_("min")=A," for "delta_("min")=,i=e`
`A=i+e-A`
`2A=i+e rArr 2A=2i rArr A=i`