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When a monochromatic light of wavelength 430 nm incident on a double slit of slit separation 5μm, there are 11 interference fringes in its central maximum. How many interference fringes will be in the central maximum of a light of the same wavelength and slit widths, but a new slit separation of 4μm?

User Yuyutsu
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Final answer:

The number of interference fringes in the central maximum of a monochromatic light incident on a double slit will change inversely with the slit separation. With the new slit separation of 4 μm, there will be 13 interference fringes in the central maximum.

Step-by-step explanation:

When a monochromatic light of wavelength 430 nm is incident on a double slit with a separation of 5 μm, resulting in 11 interference fringes in its central maximum, the number of fringes observed can be related to the slit separation. As the slit separation decreases, the spacing between interference fringes increases, meaning there would be fewer fringes within the same central maximum area.

Using the principle that the number of fringes is inversely proportional to the slit separation, if we calculate the new number of fringes for a slit separation of 4 μm, we would expect:

New Fringes = Old Fringes × (Old Separation / New Separation)

New Fringes = 11 × (5 μm / 4 μm)

New Fringes = 11 × 1.25 = 13.75

Since we can't have a fraction of a fringe, there would be 13 interference fringes in the central maximum with the new slit separation of 4 μm.

User Jawara
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