Question

When a monochromatic light of wavelength 433 nm incident on a double slit of slit separation 6 µm, there are 5 interference fringes in its central maximum. How many interference fringes will be in the central maximum of a light of wavelength 632.9 nm for the same double slit?

Answer 1

The number of interference fringes in the central maximum of a double slit can be calculated using the formula (slit separation)/(wavelength) * (distance between screen and double slit). For a light of wavelength 632.9 nm with the same double slit, there will be approximately 65.68 interference fringes in the central maximum.

Step-by-step explanation:

In a double-slit interference pattern, the number of interference fringes in the central maximum can be determined using the formula:

No. of fringes = (slit separation)/(wavelength) * (distance between screen and double slit)

Given that for the first case the wavelength is 433 nm, the slit separation is 6 µm, and there are 5 interference fringes in the central maximum, we can rearrange the formula to find the distance between the screen and double slit = (slit separation)/(wavelength) * (No. of fringes) = (6 µm)/(433 nm) * 5 = 69.07 cm.

Now, for the second case, where the wavelength is 632.9 nm and the same double slit is used, we can plug in the values into the formula: (slit separation)/(wavelength) * (distance between screen and double slit) = (6 µm)/(632.9 nm) * (69.07 cm) = 65.68 cm.

Therefore, for a light of wavelength 632.9 nm, there will be approximately 65.68 interference fringes in the central maximum of the double slit.

Answer 2

The number of interference fringes is
`n = 3`

Step-by-step explanation:

From the question we are told that

The wavelength is
`\lambda = 433 \ nm = 433 *10^(-9) \ m`

The distance of separation is
`d = 6 \mu m = 6 *10^(-6) \ m`

The order of maxima is m = 5

The condition for constructive interference is

`d sin \theta = n \lambda`

=>
`\theta = sin^(-1) [(5 * 433 *10^(-9))/( 6 *10^(-6)) ]`

=>
`\theta = 21.16^o`

So at

`\lambda_1 = 632.9 nm = 632.9*10^(-9) \ m`

`6 * 10^(-6) * sin (21.16) = n * 632.9 *10^(-9)`

=>
`n = 3`