In a double-slit experiment, the fourth-order maximum for wavelength of 450 nm occurs at an angle theta = 90. Thus it is on the verge of being eliminated from the pattern because theta cannot exceed 90. (a) What range of wavelengths in the visible range (400-700 nm) are not present in the third-order maxima?
To determine the range of wavelengths in the visible range that are not present in the third-order maxima, we first need to understand the relationship between the angle theta and the order of the maxima in a double-slit experiment.
In general, for the double-slit interference pattern, the angle theta (measured from the central maximum) for the mth order maximum is given by the equation:
sin(theta) = m * (lambda / d)
Where:
- theta is the angle of the maxima
- m is the order of the maxima
- lambda is the wavelength of light
- d is the separation between the slits
In this case, we are given that the fourth-order maximum occurs at theta = 90 degrees. Using this information, we can solve for the separation between the slits (d) using the given wavelength (lambda).
Rearranging the equation, we have:
sin(theta) = m * (lambda / d)
d = m * (lambda / sin(theta))
Plugging in the values:
m = 4
lambda = 450 nm (or 0.45 micrometers)
theta = 90 degrees
d = 4 * (0.45 x 10^-6 m / sin(90 degrees))
Now, we can calculate the separation between the slits (d):
d = 4 * (0.45 x 10^-6 m / 1)
d = 1.8 x 10^-6 m
Next, we want to find the range of wavelengths in the visible range (400-700 nm) that are not present in the third-order maxima.
For the third-order maximum, the value of m is 3. To find the corresponding wavelength range, we can rearrange the equation:
wavelength = d * sin(theta) / m
Let's calculate the upper and lower limits of this range.
Lower wavelength limit:
wavelength_lower = d * sin(90 degrees) / 3
wavelength_lower = (1.8 x 10^-6 m) * 1 / 3
wavelength_lower = 0.6 x 10^-6 m = 600 nm
Upper wavelength limit:
wavelength_upper = d * sin(89.99 degrees) / 3
wavelength_upper ≈ (1.8 x 10^-6 m) * 0.99999996 / 3
wavelength_upper ≈ 0.6 x 10^-6 m ≈ 600 nm
Therefore, the range of wavelengths in the visible range (400-700 nm) that are not present in the third-order maxima is approximately 400 nm to 600 nm.
To determine the range of wavelengths in the visible range that are not present in the third-order maxima, we need to find the wavelength range for which the angle theta exceeds 90 degrees.
In a double-slit experiment, the angle theta can be calculated using the equation:
sin(theta) = m * lambda / d,
where:
- theta represents the angle at which the maximum occurs,
- m is the order of the maximum,
- lambda is the wavelength of light used, and
- d is the distance between the two slits.
In this case, we are considering the third-order maxima, so m = 3.
We need to find the range of wavelengths for which theta exceeds 90 degrees, meaning sin(theta) > 1.
Substituting the given values into the equation, we have:
sin(theta) = (3 * lambda) / d
Since sin(theta) cannot exceed 1, we can set up the following inequality:
(3 * lambda) / d ≤ 1
Rearranging the inequality, we get:
lambda ≤ d / 3
Now, we know that the distance between the slits (d) is not provided in the question. Therefore, we cannot determine the exact values of the excluded wavelengths. However, we can conclude that any wavelength (lambda) in the visible range (400-700 nm) that satisfies the inequality lambda ≤ (d / 3) will be excluded from the third-order maxima pattern.