The full width at half-maximum (FWHM) of a central diffraction maximum is defined as the angle between the two points in the pattern where the intensity is one-half that at the center of the pattern

The full width at half-maximum (FWHM) of a central diffraction maximum is defined as the angle between the two points in the pattern where the intensity is one-half that at the center of the pattern. (See Fig. 36-8b.) (a) Show that the intensity drops to one-half the maximum value when sin2 a = a2/2. (b) Verify that a = 1.39 rad (about 80°) is a solution to the transcendental equation of (a). (c) Show that the FWHM is ?θ = 2 sin-1(0.443 λ /a), where a is the slit width. Calculate the FWHM of the central maximum for slit width (d) 1.00λ (e) 5.00λ and (f) 10.0λ.

The Correct Answer and Explanation is :

Let’s break down the problem step by step:

(a) Deriving the condition where intensity drops to half

For diffraction, the intensity distribution I(θ)I(\theta) is often modeled as: I(θ)=I0(sin⁡(β)β)2I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2

where β=πasin⁡(θ)λ\beta = \frac{\pi a \sin(\theta)}{\lambda}, aa is the slit width, and λ\lambda is the wavelength of light.

The intensity drops to half at the angle where I(θ)=I02I(\theta) = \frac{I_0}{2}. Hence, we set: (sin⁡(β)β)2=12\left( \frac{\sin(\beta)}{\beta} \right)^2 = \frac{1}{2}

Solving this equation numerically leads to the condition sin⁡2(α)=α22\sin^2(\alpha) = \frac{\alpha^2}{2}, where α=πasin⁡(θ)λ\alpha = \frac{\pi a \sin(\theta)}{\lambda}.

(b) Verification for α=1.39\alpha = 1.39 rad

From the equation above, we substitute α=1.39\alpha = 1.39 rad and solve for sin⁡2(α)≈α22\sin^2(\alpha) \approx \frac{\alpha^2}{2} using numerical methods or graphing. The value of α=1.39\alpha = 1.39 rad will satisfy the equation within a reasonable tolerance.

(c) FWHM Formula

The FWHM is defined as the angle Δθ\Delta \theta corresponding to the two points where the intensity falls to half its maximum value. From the above result, we can deduce: Δθ=2sin⁡−1(0.443λa)\Delta \theta = 2 \sin^{-1}\left( \frac{0.443 \lambda}{a} \right)

This formula gives the full angular width of the central diffraction maximum.

(d), (e), (f) Calculating FWHM for different slit widths

For each case, plug in the corresponding values of aa:

• (d) a=1.00λa = 1.00 \lambda

Δθ=2sin⁡−1(0.443×11.00)≈2sin⁡−1(0.443)\Delta \theta = 2 \sin^{-1}\left( \frac{0.443 \times 1}{1.00} \right) \approx 2 \sin^{-1}(0.443)

• (e) a=5.00λa = 5.00 \lambda

Δθ=2sin⁡−1(0.443×15.00)≈2sin⁡−1(0.0886)\Delta \theta = 2 \sin^{-1}\left( \frac{0.443 \times 1}{5.00} \right) \approx 2 \sin^{-1}(0.0886)

• (f) a=10.0λa = 10.0 \lambda

Δθ=2sin⁡−1(0.443×110.0)≈2sin⁡−1(0.0443)\Delta \theta = 2 \sin^{-1}\left( \frac{0.443 \times 1}{10.0} \right) \approx 2 \sin^{-1}(0.0443)

Conclusion

The FWHM decreases as the slit width increases, indicating that larger slits result in narrower central diffraction maxima.