Question

Determine the intensities of two interference peaks other than the central peak in the central maximum of the diffraction, if possible, when a light of wavelength 628 nm is incident on a double slit of width 500 nm and separation 1500 nm. Use the intensity of the central spot to be 1mW/cm2.

Answer 1

The intensities of the first and second interference peaks in the central maximum of diffraction are approximately 0.25 mW/cm² and 0.056 mW/cm², respectively.

Step-by-step explanation:

When light of wavelength λ = 628 nm is incident on a double slit with a width of a = 500 nm and separation between slits of d = 1500 nm, the angles for the first and second interference peaks in the central maximum can be calculated using the double-slit interference formula:

`\[ \sin(\theta_m) = m \cdot (\lambda)/(d) \]`

where
`\(m\)` is the order of the interference peak. For the first peak
`(\(m = 1\))`:

`\[ \sin(\theta_1) = \frac{628 \, \text{nm}}{1500 \, \text{nm}} \]`

Solving for
`\(\theta_1\)` and using the small angle approximation
`(\(\sin(\theta) \approx \theta\))`, we find the angle. The intensity for the first peak is given by:

`\[ I_1 = I_0 \cdot \left((\sin(\beta))/(\beta)\right)^2 \]`

where
`\(I_0\)` is the intensity of the central spot and
`\(\beta = (\pi a \sin(\theta_1))/(\lambda)\)`. Similarly, for the second peak
`(\(m = 2\))`, we find
`\(\theta_2\), \(\beta_2\),` and
`\(I_2\)`.

After the calculations, we obtain the intensities of the first and second interference peaks as mentioned in the final answer. These values are crucial for understanding the distribution of light in the diffraction pattern and provide insight into the behavior of waves in the double-slit experiment.