The correct answer is 2
\(I_{R_1}=I_1+I_2+2\sqrt{I_1I_2}\cos\phi\)
\(I_A=I+4I+2\sqrt{I⋅4I}\cos90^∘=5I\)
\(I_B=I+4I+2\sqrt{I⋅4I}\cos60^∘=7I\)
\(I_B–I_A=2I\)
\(\therefore\) value of x is 2
Calculate the angle of minimum deviation of an equilateral prism. The refractive index of the prism is \(\sqrt{3}\). Calculate the angle of incidence for this case of minimum deviation also.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: