question_answer

A ray of light of intensity I is incident on a parallel glass-slab at a point A as shown in fig. It undergoes partial reflection and refraction. At each reflection 25% of incident energy is reflected. The rays AB and A'B' undergo interference. The ratio \[{{I}_{\max }}/{{I}_{\min }}\] is                                                                [IIT 1990]

A)            4 : 1                                          

B)            8 : 1

C)            7 : 1                                          

D)            49 : 1

Correct Answer: D

Solution :

             From figure \[{{I}_{1}}=\frac{I}{4}\] and \[{{I}_{2}}=\frac{9I}{64}\]  Þ \[\frac{{{I}_{2}}}{{{I}_{1}}}=\frac{9}{16}\] By using \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\left( \frac{\sqrt{\frac{{{I}_{2}}}{{{I}_{1}}}}+1}{\sqrt{\frac{{{I}_{2}}}{{{I}_{1}}}}-1} \right)=\left( \frac{\sqrt{\frac{9}{16}}+1}{\sqrt{\frac{9}{16}}-1} \right)=\frac{49}{1}\]