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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 $$5$$th bright fringe at the same point?


A
630 nm
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B
500 nm
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C
420 nm
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D
750 nm
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Solution

The correct option is A $$420\ nm$$
Position of nth bright fringe  $$y_n    = \dfrac{n\lambda  D}{d}$$
Given : $$\lambda  = 700\ nm$$

$$\therefore$$   $$y_3  = \dfrac{3  \times 700  D}{d}$$ $$\implies   y_3  = \dfrac{2100  D}{d} $$

The 5th bright fringe due to light of wavelength $$\lambda'$$ is formed at   $$y_3$$
$$\therefore$$    $$y_5  =y_3 $$

OR $$\dfrac{5 \times \lambda' D}{d}  = \dfrac{2100   D}{d}$$

$$\implies$$   $$\lambda' = \dfrac{2100  }{5}  =420\ nm$$

Physics

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