Question

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Q.In an interference experiment third bright fringes obtained at a point on the screen in the light of 700 nm what should be the wavelength of the light source in order to obtain 5th bright fringe at the same. ​

Answer 1

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using n1 lemda 1 = n2 lemda 2

we get

3×700= 5× lemda 2

lemda 2 = 3×700/5

= 420nm

Answer 2

Explanation:

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n1 lemda 1 =n2 lemda 2

we get

3×700=5×lemda 2

lemda 2=3×700/5

=420 nm