Question

Q37 When a beam of white light is incident at
an angle of 30° on a glass slab, the
distance between red and violet rays as
observed on a screen on the other side
of the slab is 0.1 mm. Calculate the
approximate width of the glass slab.
(Note: Use violet = 1.66 and pred = 1.6)

Answer 1

Given:

• Glancing angle $= 20 ^\circ$

• $\mu_{violet} = 1.66$

• $\mu_{red} = 1.6$

To Find:

• The  approximate width of the glass slab.

Solution:

Glancing angle $= 20 ^\circ$

Therefore, angle of incidence $= 90 ^\circ - 20 ^\circ = 70 ^\circ$

            $sinr_v = \frac{70 ^\circ}{1.66}$

             $r_v = 34^\circ 30\minute$

           $sinr_r = \frac{70 ^\circ}{1.6}$

          $r_r = 36^\circ$

we know that,

           $l= \frac{tsin(i-r)}{cosr}$

   $l_v-l_ r = d[\frac{sin(i-r_v)}{cosr_v} - \frac{sin(i-r_r)}{cosr_r}]$

        $0.1 = d[\frac{sin 35^\circ 30}{cos 34^\circ 30} - \frac{sin 34^\circ}{cos 36^\circ}]$

        $0.1 = d[\frac{0.5807}{0.8241} - \frac{55.92}{0.8090}]$

        $0.1 = d[0.71-0.68]$

           $d = \frac{10}{3} cm$

Thus, The  approximate width of the glass slab $d = \frac{10}{3} cm$