question_answer

Three right angled prisms of refractive indices \[{{n}_{1}},{{n}_{2}}\] and \[{{n}_{3}}\] are fixed together using an optical glue as shown in figure. If a ray passes through the prisms without suffering any deviation, then

A)            \[{{n}_{1}}={{n}_{2}}={{n}_{3}}\]                                    

B)            \[{{n}_{1}}={{n}_{2}}\ne {{n}_{3}}\]

C)            \[1+{{n}_{1}}={{n}_{2}}+{{n}_{3}}\]                               

D)            \[1+n_{2}^{2}=n_{1}^{2}+n_{3}^{2}\]

Correct Answer: D

Solution :

At B \[\sin i={{n}_{1}}\sin {{r}_{1}}\] Þ \[{{\sin }^{2}}i=n_{1}^{2}{{\sin }^{2}}{{r}_{1}}\]                      .... (i) At C \[{{n}_{1}}\sin (90-{{r}_{1}})={{n}_{2}}\sin {{r}_{2}}\]Þ\[n_{2}^{2}{{\sin }^{2}}{{r}_{2}}=n_{1}^{2}{{\cos }^{2}}{{r}_{1}}\]....(ii) At D \[{{n}_{2}}\sin (90-{{r}_{2}})={{n}_{3}}\sin {{r}_{3}}\]Þ\[n_{2}^{2}{{\cos }^{2}}{{r}_{2}}=n_{3}^{2}{{\sin }^{2}}{{r}_{3}}\]           ....(iii) At E \[{{n}_{3}}\sin (90-{{r}_{3}})=(1)\sin (90-1)\]Þ\[{{\cos }^{2}}i=n_{3}^{2}{{\cos }^{2}}{{r}_{3}}\]  ....(iv) Adding (i), (ii), (iii) and (iv) we get \[1+n_{2}^{2}=n_{1}^{2}+n_{3}^{2}\]