Three right angled prisms of refractive indices $n_1,n_2\text{ and }{n}_3$ are fixed together using an optical glue as shown in fig. If a ray passes through the prisms without suffering any deviation, then

$\text{ At B }\mathrm{Sin}i=n_1\sin r_1$

${\sin }^{2}i=n_1^2{\sin }^2{r}_1\to \left(1\right)$

At C

$n_1\sin \left(90-r_1\right)=n_2\sin r_2$

$n_2^2{\sin }^2{r}_2=n_1^2{\cos }^2{r}_1\to \text{ (2) }$

At D

$n_2\sin \left(90-r_2\right)=n_3\sin r_3$

$n_2^2{\cos }^2{r}_2=n_3^2{\sin }^2{r}_3\to \left(3\right)$

At E

$n_3\sin \left(90-r_3\right)=\left(1\right)\sin (90-i)$

$n_3^2{\cos }^2{r}_3={\cos }^{2}i\to \left(4\right)$

Adding (1),(2), (3) & (4)

$1+n_2^2=n_1^2+n_3^2$