A Fraunhofer diffration pattern is produced on a screen 210cm from a single slit. The distance from the center of the central maximum to the first-order maximum is 7560 times the wavelength (lambda). Calculate slit width. Answer in __m
To solve this problem, we can use the formula for the angular position of the first-order maximum in a single slit diffraction pattern:
sin(θ) = λ / (w)
Where:
- θ is the angular position of the first-order maximum,
- λ (lambda) is the wavelength of the light,
- w is the width of the slit.
From the given information, we know that the distance from the center of the central maximum to the first-order maximum is 7560 times the wavelength (λ). In other words:
d = 7560λ
We can also determine the angular position of the first-order maximum using trigonometry. Given that the distance from the screen to the single slit is 210 cm, we have a right triangle with:
opposite side = d (7560λ)
adjacent side = distance to the screen (210 cm)
Using the inverse sine function, we can find the angle θ:
θ = arcsin(d / distance)
θ = arcsin((7560λ) / 210)
Now, we can substitute this value of θ into the first equation:
sin(θ) = λ / (w)
Rearranging this equation, we can solve for the width of the slit, w:
w = λ / sin(θ)
w = λ / sin(arcsin((7560λ) / 210))
Simplifying further:
w = λ / ((7560λ) / 210)
w = 210 / 7560
w = 0.02777778 m
Therefore, the slit width is approximately 0.028 m or 2.8 cm.