The location of the first dark band of the single-slit diffraction pattern is given by the formula:
\[ y = \frac{(m + \frac{1}{2})\lambda L}{a} \]
Where:
\( y = 4 \times 10^{-2} \mathrm{m} \) (distance to the first dark band)
\( m = 0 \) (for the first dark band)
\( \lambda \) is the wavelength of the light
\( L = 50 \mathrm{cm} = 0.5 \mathrm{m} \) (distance to the screen)
\( a = 1.2 \times 10^{-7} \mathrm{m} \) (width of the slit)
Substitute the given values into the formula:
\[ 4 \times 10^{-2} = \frac{(0 + \frac{1}{2}) \lambda (0.5)}{1.2 \times 10^{-7}} \]
\[ 4 \times 10^{-2} = \frac{0.5 \lambda}{1.2 \times 10^{-7}} \]
\[ \lambda = \frac{4 \times 10^{-2} \times 1.2 \times 10^{-7}}{0.5} = 9.6 \times 10^{-7} \mathrm{m} \]
Therefore, the wavelength of the light is \( 9.6 \times 10^{-7} \mathrm{~m} \).
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The momentum of a photon is given by the formula:
\[ p = \frac{h}{\lambda} \]
Where:
\( h = 6.63 \times 10^{-34} \mathrm{~J\cdot s} \) (Planck's constant)
\( \lambda = 1.4 \times 10^{-7} \mathrm{~m} \) (wavelength of the light)
Substitute the values into the formula:
\[ p = \frac{6.63 \times 10^{-34}}{1.4 \times 10^{-7}} = 4.7 \times 10^{-27} \mathrm{~kg\cdot m/s} \]
Therefore, the momentum of a photon with a wavelength of \( 1.4 \times 10^{-7} \mathrm{~m} \) is \( 4.7 \times 10^{-27} \mathrm{~kg\cdot m/s} \).
11. Light is passed through a single slit that has a width of \( 1.2 \times 10^{-7} \mathrm{~m} \). What is the wavelength of the light if the first dark band is \( 4 \mathrm{~cm} \) from the central bright on a screen \( 50 \mathrm{~cm} \) away? [ \( 9.6 \times 10^{-} \mathrm{m} \) ]
What is the momentum of a photon with a wavelength of \( 1.4 \times 10^{-7} \mathrm{~m} \) ? \( \left[4.7 \times 10^{27} \mathrm{~kg} \mathrm{~m} / \mathrm{s}\right] \)
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