Figure.. Shows a triangular prism of refracting angle `90^(@)` . A ray of light incident at face AB at an angle `theta _(1)`, refracts at point Q with an angle of refraction `90^(@)`.
a. What is the regractive index of the prism in terms of `theta_(1)`?
b. What is the maximum value that the refractive inde can have?
c. What happens to the light at Q if the incident angle at Q is (i) increased slightly, and (ii) decrease slightly?

Explanation

a. Let the ray be incident at an angle `theta_(1)` at the face AB. It refracts at an angle `theta_(2)` and is incident at an angle `theta_(3)` at the face AC. Finally, the ray comes out at an angle `theta_(4)=90^@`.
From Fig. , the normal at faces AB and AB make an angle of `90^@` width each other, `theta_(3)=90^@-theta_(2)`
`sintheta_(3)=sin(90^(@)-theta_(2))=costheta_(2)=sqrt(1-sin^(2)theta_(2))` (i)
From Snell's law at face AC, `n sintheta_(2)=1`
`sqrt(1-sin^(2)theta_(2))=1` (ii)
Form Snell's law at face AB, `1sintheta_(1)=nsintheta_(2)`
`sintheta_(2)=(sing theta_(1))/(n)` (iii)
From Eqs. (ii) and (iii), we have
`nsqrt(1-(sin^(2)theta_(1))/(n^(2)))=1` (iv)
On squaring Eq. (iv) and solving for n, we get
`n=sqrt(1+(sin^(2)theta_(1)))`
b. The greatest possible value of `sin^(2)theta_(1)` is 1, hence, the greates possible value of n is `n_(max)=sqrt(2)=1.41`.
c. i. For a given n, if `theta_(1)` is increased the angle of refraction `theta_(2)` increases. As `theta_(3)=90^@-theta_(2)` the angle `theta` decreases, i.e., the angle of incidence at face AC is less than the critical angle for total reflection, hence light light emerges into air.
ii. If the angle of incidence is decresed, the angle of refraction `theta_(2)` decreases. So, the angle `theta_(3)` increases. The angle of incidence at the second surface is greater than the critical angle, so light is reflected at Q.