Solve for x. (a + b)2x2 – 8(a2 – b2)x – 20(a – b)2 = 0
Answers
Answer:
Given :-
The distance between two consecutive bright fringes in a baptism experiment using light of wavelength 6000 A⁰is 0.32mm
To Find :-
how much will the distance change if light of wavelength 4800A⁰is used?
Solution :-
We know that
\sf \beta _{i} \propto \lambda_{i}\; and \; \beta_{f}\propto \lambda_{f}βi∝λiandβf∝λf
\sf \dfrac{0.32\times 10^{-3}}{\beta_{f}} = \dfrac{6000 \times 10^{-10}}{4800 \times 10^{-10}}βf0.32×10−3=4800×10−106000×10−10
\sf \dfrac{0.32\times10^{-3}}{\beta_{f}} = \dfrac{6000}{4800}βf0.32×10−3=48006000
\sf\dfrac{0.32\times 10^{-3}}{\beta_{f}} = \dfrac{60}{48}βf0.32×10−3=4860
\sf\dfrac{0.32\times 10^{-3}}{\beta_{f}} = 1.25βf0.32×10−3=1.25
\sf 0.32\times 10^{-3} = 1.25\times\beta_{f}0.32×10−3=1.25×βf
\sf \dfrac{0.32\times 10^{-3}}{1.25}=\beta_{f}1.250.32×10−3=βf
\sf 0.25 \times 10^{-3}=\beta_{f}0.25×10−3=βf
Now
\sf Difference = 0.32 \times 10^{-3}-0.25\times 10^{-3}Difference=0.32×10−3−0.25×10−3
\sf Difference = 0.7\times 10^{-3}\;mDifference=0.7×10−3m
Answer:
Solve for x. (a + b)2x2 – 8(a2 – b2)x – 20(a – b)2 = 0
