Question

Let $r$ and ${r^{'}}$ denote the angle inside an equilateral prism, as usual, in degrees. Consider that during some time interval from $t = 0$ to $t$ , ${r^{'}}$ varies with time as ${r^{'}} = 10 + {t^{2}}$. During this time, $r$ will vary as
A. $50 - {t^{2}}$
B. $50 + {t^{2}}$
C. $60 - {t^{2}}$
D. $60 + {t^{2}}$

Answer 1

Hint: In this question we will derive a relation between $r$ and ${r^{'}}$ with the angle of prism by using simple properties of prism. The angle made in equilateral prism $A = {60^ \circ }$. We will put the values of ${r^{'}}$ in terms of time and equate them to get $r$.

Formula used:
Angle of prism=
$A = {r^{'}} + r$

Complete step by step solution:

Given in the diagram an incident ray enters the prism and makes angle $r$ with perpendicular line from one of its faces and an incident on another face at angle ${r^{'}}$with respect to perpendicular on that face.
So, from the property of triangle, in a prism $A = {r^{'}} + r$
Given that ${r^{'}}$ varies with time from time equal to zero to ${r^{'}} = 10 + {t^{2}}$
$ \Rightarrow {60^ \circ } = 10 + {t^{2}} + r$
As given it is an equilateral prism,
$ \Rightarrow r = 50 - {t^{2}}$
So, during the time ${r^{'}}$ varies with time from time equal to zero to ${r^{'}} = 10 + {t^{2}}$ , $r$ will vary as $r = 50 - {t^{2}}$ .
Hence option A is correct.

Note: The prism's angle is formed by its two sides or surfaces, through which incident light enters and emerging light escapes. As it is an equilateral prism, the prism's angles will be equal to ${60^ \circ }$ . Care should be taken that using the equation all the angles should be in degree.