A) x + y = 1
B) x ? y = 1
C) x + y = - 1
D) x ? y = - 1
Correct Answer: A
Solution :
Given curve is \[=\frac{{{10}^{5}}}{2\pi }Hz\] On differentiating w. R. t. X, we get \[E=\frac{12375}{5000}\] \[=2.475eV=4\times {{10}^{-19}}J\] \[{{\text{l}}_{\text{eye}}}\text{=}\left( \text{photon flux} \right)\text{ }\!\!\times\!\!\text{ energy of a photon}\] \[{{\text{l}}_{\text{eye}}}\text{=}\left( 5\times {{10}^{4}} \right)\times 4\times {{10}^{-19}}\] [ \[=2\times {{10}^{-14}}\left( W/{{m}^{2}} \right)\] at x = 0, y = 1] Slope of normal at (x = 0) = - 1 \[=\frac{{{l}_{ear}}}{{{l}_{eye}}}=\frac{{{10}^{-13}}}{2\times {{10}^{-14}}}=5\] Equation of normal at x = 0 and y = 1 is Y ? 1 = - 1 (x ? 0) \[\mu =\frac{\Delta {{V}_{p}}}{\Delta {{V}_{g}}}\] y ? 1 = - x \[\Rightarrow \] x + y = 1
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