a screen is placed at a distance of 1.5 m from a narrow slit. the slit is illuminated by light of wavelength 6000 angstrom if the first minimum on either side of the central maximum is at a distance of 5 mm from , find the width of the slit
Well, well, well, let's shed some light on this problem, shall we?
To find the width of the slit, we can use the concept of diffraction. The formula for the location of the first minimum in the diffraction pattern surrounding a narrow slit is given by:
sin(θ) = λ / (w * n)
Where:
- θ is the angle between the central maximum and the first minimum,
- λ is the wavelength of light (6000 Å = 6000 * 10^(-10) m),
- w is the width of the slit, and
- n is the order of the minimum (in this case, n = 1).
Now, let's convert all the units and plug in the values:
sin(θ) = 6000 * 10^(-10) m / (w * 1)
Since we want to find the width of the slit, we rearrange the equation:
w = 6000 * 10^(-10) m / sin(θ)
Now, you mentioned that the first minimum is at a distance of 5 mm from the central maximum. We can use some trigonometry to relate this distance to the angle θ.
We have a right-angled triangle where the opposite side is the distance of 5 mm (0.005 m) and the hypotenuse is the distance between the screen and the slit, which is 1.5 m.
sin(θ) = opposite / hypotenuse
sin(θ) = 0.005 / 1.5
Now we can plug this value into our equation for the slit width:
w = 6000 * 10^(-10) m / sin(θ)
w = 6000 * 10^(-10) m / (0.005 / 1.5)
Time to whip out the calculator!
w ≈ 0.12 * 10^(-3) m
So, dear reader, the width of the slit is approximately 0.12 millimeters. Remember, this answer came with a dash of humor, courtesy of your friendly neighborhood Clown Bot!
To find the width of the slit, we can use the concept of diffraction.
The first minimum in a single-slit diffraction pattern occurs at an angle θ given by:
sin(θ) = λ / (a * n),
where λ is the wavelength of light, a is the width of the slit, and n is the order of the minimum.
In this case, the first minimum on either side of the central maximum occurs at a distance of 5 mm from the central maximum. Since the distance is small, we can express it as an angle in radians:
θ = (5 mm) / (1.5 m)
Next, we need to convert the wavelength from angstroms to meters:
λ = 6000 angstrom = 6000 × 10^(-10) m
Now we can substitute the values into the equation to find the width of the slit:
sin(θ) = λ / (a * n)
sin((5 mm) / (1.5 m)) = (6000 × 10^(-10) m) / (a * 1)
Rearranging the equation, we can solve for a:
a = (6000 × 10^(-10) m) / (sin((5 mm) / (1.5 m)))
Calculating this equation will give you the width of the slit in meters.
To find the width of the slit, we can use the concept of diffraction and the formula for the position of the first minimum.
The formula for the position of the first minimum in a single slit diffraction pattern is given by:
sin(theta) = λ / (w * d)
Where:
- theta is the angle between the central maximum and the first minimum,
- λ is the wavelength of light,
- w is the width of the slit,
- d is the distance between the slit and the screen.
Here, we are given:
- d = 1.5 m,
- λ = 6000 Å = 6000 * 10^(-10) m,
- The distance between the central maximum and the first minimum (sin(theta)) is 5 mm = 5 * 10^(-3) m.
Now, we can rearrange the formula to solve for w:
w = λ / (sin(theta) * d)
Substituting the known values:
w = (6000 * 10^(-10) m) / (5 * 10^(-3) m * 1.5 m)
Simplifying the expression:
w = (6000 * 10^(-10)) / (5 * 10^(-3) * 1.5)
w = 0.008 mm
Therefore, the width of the slit is approximately 0.008 mm.