You illuminate a slit with a width of 70.5 um with a light of wavelength 727 nm and observe the resulting diffraction pattern on a screen that is situated 2.83 m from the slit. What is the width, in centimeters, of the pattern’s central maximum? width: 8.8 cm Incorrect Light of wavelength 558.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 86.5 cm from the slit. The distance on the screen between the third order minimum and the central maximum is 1.87 cm. What is the width a of the slit in micrometers (um)? a= 5.16 x10-5 um Incorrect
The Correct Answer and Explanation is :
To solve these diffraction problems, we need to use the diffraction formula for a single slit, which is given by:
[
\sin(\theta_m) = \frac{m \lambda}{a}
]
where:
- ( \theta_m ) is the angle for the m-th order minimum (the first minimum corresponds to ( m = 1 )),
- ( m ) is the order of the minimum (integer),
- ( \lambda ) is the wavelength of the light,
- ( a ) is the slit width.
Problem 1: Width of the Central Maximum
In this case, the diffraction pattern’s central maximum is the region between the first minimum on either side of the central maximum. For a single-slit diffraction pattern, the angle ( \theta_1 ) for the first minimum (m = 1) is:
[
\sin(\theta_1) = \frac{\lambda}{a}
]
To calculate the width of the central maximum, we need to find the distance between the two first-order minima. If ( L ) is the distance from the slit to the screen and ( y_1 ) is the distance from the central maximum to the first minimum on either side, the following relationship holds:
[
y_1 = L \tan(\theta_1)
]
For small angles, ( \tan(\theta_1) \approx \sin(\theta_1) ), so:
[
y_1 = L \frac{\lambda}{a}
]
Thus, the total width of the central maximum (from the first minimum on one side to the first minimum on the other side) is:
[
\text{Width} = 2y_1 = 2L \frac{\lambda}{a}
]
Plugging in the values:
- ( \lambda = 727 \, \text{nm} = 727 \times 10^{-9} \, \text{m} ),
- ( a = 70.5 \, \mu\text{m} = 70.5 \times 10^{-6} \, \text{m} ),
- ( L = 2.83 \, \text{m} ),
[
\text{Width} = 2 \times 2.83 \times \frac{727 \times 10^{-9}}{70.5 \times 10^{-6}} = 0.088 \, \text{m} = 8.8 \, \text{cm}
]
Thus, the width of the central maximum is 8.8 cm, which matches the given answer.
Problem 2: Slit Width
In this case, we are given the distance between the third-order minimum and the central maximum, which is 1.87 cm. The distance for the third-order minimum, ( y_3 ), is given by:
[
y_3 = L \frac{3 \lambda}{a}
]
From the given values:
- ( L = 86.5 \, \text{cm} = 0.865 \, \text{m} ),
- ( y_3 = 1.87 \, \text{cm} = 0.0187 \, \text{m} ),
- ( \lambda = 558.0 \, \text{nm} = 558 \times 10^{-9} \, \text{m} ).
Rearranging the formula for ( a ):
[
a = \frac{3 \lambda L}{y_3}
]
Substitute the values:
[
a = \frac{3 \times 558 \times 10^{-9} \times 0.865}{0.0187}
]
[
a \approx 5.16 \times 10^{-5} \, \text{m} = 51.6 \, \mu\text{m}
]
Thus, the slit width is approximately ( 51.6 \, \mu\text{m} ).
This solution method relies on applying the diffraction equations and knowing how to relate the measured distances to the geometry of the diffraction pattern.