**(a)**

Width of central maximum for single slit:

$\overline{){\mathbf{w}}{\mathbf{=}}\frac{\mathbf{2}\mathbf{L}\mathbf{\lambda}}{\mathbf{a}}}$

The width of the central maximum in the intensity pattern for a single slit is given by 1. w = 2Lλ/a

(Figure 1) shows the light intensity on a viewing screen behind a single slit of width a. The light's wavelength is λ.

Is λ < a, λ = a, λ > a, or is it not possible to tell?

Answer the question by filling in the terms or equations to the appropriate blanks to complete the sentences.

(a) The width of the central maximum in the intensity pattern for a single slit is given by ___________.

(b) If λ ≥ a than w will be equal to or ________ 2L.

(c) It is obvious that w << 2L, hence λ is _______ a.

- w = 2Lλ/a
- w = 2aL/λ
- w = 2La/λ
- greater than
- less than
- equal to
- w = 2aλ/L

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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Single Slit Diffraction concept. You can view video lessons to learn Single Slit Diffraction. Or if you need more Single Slit Diffraction practice, you can also practice Single Slit Diffraction practice problems.

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Based on our data, we think this problem is relevant for Professor Eberle's class at NORTHLAKECOLLEGE.