Question
In an interference experiment, third bright fringe is obtained at a point on the screen with a light of $700 \,nm$. What should be the wavelength of the light source in order to obtain 5th bright fringe at the same point......$nm$
✓
Answer
d
(d)${n_1}{\lambda _1} = {n_2}{\lambda _2} \Rightarrow 3 \times 700 = 5 \times {\lambda _2} \Rightarrow {\lambda _2} = 420\;nm$
(d)${n_1}{\lambda _1} = {n_2}{\lambda _2} \Rightarrow 3 \times 700 = 5 \times {\lambda _2} \Rightarrow {\lambda _2} = 420\;nm$
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