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 centre of the pattern, as shown in the Figure (b).
part 1 YSE/FALSE:
a.Does the intensity drop to one-half the maximum value when sin2? = ?2/2?
b.Is ? = 1.39 rad (about 80°) a solution to the transcendental equation of the previous question?
c.Is the FWHM ?? = 2 sin-1(0.442 ?/a), where a is the slit width?
part 2:
a.Calculate the FWHM of the central maximum for slit width 1.66 ?
b.Calculate the FWHM for 5.58 ?
c.Calculate the FWHM For 10.86 ?.
The Correct Answer and Explanation is :
Part 1: True/False
a. Does the intensity drop to one-half the maximum value when sin²θ = λ/2a?
False.
For a single-slit diffraction pattern, the intensity I(θ) as a function of angle θ is given by:
[ I(\theta) = I_0 \left(\frac{\sin(\beta)}{\beta}\right)^2 ]
where (\beta = \frac{\pi a}{\lambda} \sin(\theta)), (a) is the slit width, and (\lambda) is the wavelength. The condition for the intensity to drop to half the maximum value occurs when (\sin^2(\theta) \approx 0.442 \lambda/a), not when (\sin^2(\theta) = \lambda/2a). The intensity doesn’t drop to exactly half at that value.
b. Is θ = 1.39 rad (about 80°) a solution to the transcendental equation of the previous question?
False.
To find the angle where the intensity drops to half of its maximum, we use the relation:
[ \frac{\sin(\beta)}{\beta} = \frac{1}{\sqrt{2}} ]
Solving this equation requires numerical methods and is not as simple as assuming (\theta = 1.39) rad. Therefore, 1.39 rad is not a straightforward solution.
c. Is the FWHM (\Delta\theta = 2 \sin^{-1}(0.442 \lambda/a)), where a is the slit width?
True.
This is the correct formula for the Full Width at Half Maximum (FWHM) in a single-slit diffraction pattern. It gives the angular width between the two points where the intensity falls to half of its central maximum. The factor 0.442 comes from solving the diffraction equation for the angle at which the intensity drops to half its peak value.
Part 2: Calculating FWHM
The formula for FWHM of the diffraction pattern is:
[
\Delta \theta = 2 \sin^{-1}\left(0.442 \frac{\lambda}{a}\right)
]
where (\lambda) is the wavelength and (a) is the slit width.
Assuming the wavelength (\lambda = 500 \, \text{nm} = 5 \times 10^{-7} \, \text{m}), we can calculate the FWHM for different slit widths.
a. For slit width (a = 1.66 \, \mu\text{m} = 1.66 \times 10^{-6} \, \text{m}):
[
\Delta \theta = 2 \sin^{-1}\left(0.442 \frac{5 \times 10^{-7}}{1.66 \times 10^{-6}}\right)
]
[
\Delta \theta = 2 \sin^{-1}(0.133) \approx 2 \times 7.63^\circ = 15.26^\circ
]
b. For slit width (a = 5.58 \, \mu\text{m} = 5.58 \times 10^{-6} \, \text{m}):
[
\Delta \theta = 2 \sin^{-1}\left(0.442 \frac{5 \times 10^{-7}}{5.58 \times 10^{-6}}\right)
]
[
\Delta \theta = 2 \sin^{-1}(0.079) \approx 2 \times 4.53^\circ = 9.06^\circ
]
c. For slit width (a = 10.86 \, \mu\text{m} = 10.86 \times 10^{-6} \, \text{m}):
[
\Delta \theta = 2 \sin^{-1}\left(0.442 \frac{5 \times 10^{-7}}{10.86 \times 10^{-6}}\right)
]
[
\Delta \theta = 2 \sin^{-1}(0.0204) \approx 2 \times 1.17^\circ = 2.34^\circ
]
Explanation:
The Full Width at Half Maximum (FWHM) of a diffraction pattern is the angular width of the central diffraction maximum where the intensity falls to half of its maximum value. The formula to calculate this angle involves the slit width (a) and the wavelength of light (\lambda).
When the slit width increases, the diffraction pattern becomes narrower, meaning the central maximum gets smaller. As seen from the calculations, a larger slit width results in a smaller FWHM, which means the diffraction is less pronounced. This behavior is consistent with the wave nature of light, where diffraction effects become less significant with increasing slit width.
The calculation of FWHM is crucial in understanding the resolving power of optical systems. The smaller the FWHM, the better the system is at distinguishing between closely spaced objects.