When 450-nm light is incident normally on a certain double-slit system the number of interference maxima within the central diffraction maximum is 5. When 900-nm light is incident on the same slit system the number is: a. 9b. 2c. 3d. 10e. 5

Question

Damaris Gilmore

Physics

31 March, 07:42

When 450-nm light is incident normally on a certain double-slit system the number of interference maxima within the central diffraction maximum is 5. When 900-nm light is incident on the same slit system the number is: a. 9b. 2c. 3d. 10e. 5

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Fletcher Shields · Answer 1

width of central diffraction maxima = 2 λD / d₁

λ is wave length of light, D is screen distance and d₁ is slit width

width of each interference fringe = λD / d₂, d₂ is slit separation.

No of interference fringe in central diffraction fringe

= width of central diffraction maxima / width of each interference fringe

= 2 λD / d₁ x λDd₂ / λD

No = 2 d₂ / d₁

No = 5

5 = 2 d₂ / d₁

Since this number does not depend upon wavelength so it will remain the same

No of required fringe will be 5.

right option

e) 5.