Light containing two different wavelengths passes through a diffraction grating with 1300 slits/cm. On a screen 16.0 cm from the grating, the third-order maximum of the shorter wavelength falls midway between the central maximum and the first side maximum for the longer wavelength. If the neighboring maxima of the longer wavelength are 8.44 mm apart on the screen, what are the wavelengths in the light? (Hint: Use the small angle approximation.)

Question

Light containing two different wavelengths passes through a diffraction grating with 1300 slits/cm. On a screen 16.0 cm from the grating, the third-order maximum of the shorter wavelength falls midway between the central maximum and the first side maximum for the longer wavelength. If the neighboring maxima of the longer wavelength are 8.44 mm apart on the screen, what are the wavelengths in the light? (Hint: Use the small angle approximation.)

Answer 1

1/130000 * sinTheta=1*lambda

but sinTheta=tantheta=8.44/160

now solve for lambda, in m

Answer 2

To solve this problem, we can use the equation for the grating spacing in the small angle approximation:

dsinθ = mλ

where d is the grating spacing, θ is the angle between the incident light and the diffracted light, m is the order of the maximum, and λ is the wavelength.

Let's define the grating spacing as d = 1/1300 cm.

For simplicity, let's denote the shorter wavelength as λ_s and the longer wavelength as λ_l.

Given that the third-order maximum of the shorter wavelength falls midway between the central maximum and the first side maximum for the longer wavelength, we can set up the following equation:

d(sinθ_s) = 3λ_s (Eq. 1)

d(sinθ_l) = λ_l (Eq. 2)

We can also use the given information that the neighboring maxima of the longer wavelength are 8.44 mm apart on the screen. Since the distance from the screen to the grating is 16.0 cm, the angle between neighboring maxima can be approximated as:

θ_l ≈ (8.44 mm) / (16.0 cm) ≈ 0.5275 radians (using the small angle approximation)

Now, we substitute the values into Eq. 2:

(1/1300 cm)(sinθ_l) = λ_l

(sinθ_l) = (1300 cm)(λ_l) (Eq. 3)

For the third-order maximum of the shorter wavelength, we know that it falls midway between the central maximum and the first side maximum for the longer wavelength. This means the angle, θ_s, between the central maximum and the third-order maximum is half of the angle, θ_l, which is:

θ_s = (1/2)θ_l ≈ (1/2)(0.5275 radians) ≈ 0.26375 radians

Now, we substitute the values into Eq. 1:

(1/1300 cm)(sinθ_s) = 3λ_s

(sinθ_s) = (3900 cm)(λ_s) (Eq. 4)

By comparing Eq. 3 and Eq. 4, we find:

(1300 cm)(λ_l) = (3900 cm)(λ_s)

λ_l = 3λ_s (Eq. 5)

Now, we can solve for λ_l and λ_s by substituting Eq. 5 into Eq. 3:

(1300 cm)(3λ_s) = (3900 cm)(λ_s)

3900λ_s = 3900λ_s

Therefore, λ_l = 3 times the value of λ_s.

In conclusion, the longer wavelength of the light is three times the value of the shorter wavelength.

Answer 3

To solve this problem, we can use the concept of diffraction and the small angle approximation. The small angle approximation states that for small angles, the tangent of the angle is approximately equal to the angle itself in radians.

Let's denote the shorter wavelength as λ1 and the longer wavelength as λ2.

First, let's find the angle at which the third-order maximum of the shorter wavelength occurs. We can use the formula for the position of the m-th order maximum of a diffraction grating:

dsinθ = mλ

Here, d is the spacing between slits, θ is the angle measured from the central maximum to the m-th order maximum, and m is the order of the maximum.

For the third-order maximum of the shorter wavelength (λ1), we have:

d*sinθ1 = 3λ1

Since the angle θ1 is small, we can use the small angle approximation, which allows us to write sinθ1 ≈ θ1. Thus, we can rewrite the equation as:

dθ1 = 3λ1

Next, let's find the angle at which the first side maximum of the longer wavelength occurs. Using the same formula for the position of the m-th order maximum, we have:

d*sinθ2 = λ2

Again, applying the small angle approximation, we get:

dθ2 = λ2

Now, we are given that the third-order maximum of the shorter wavelength falls midway between the central maximum and the first side maximum of the longer wavelength. This means that the angle for the third-order maximum of the shorter wavelength (θ1) is halfway between the angle for the central maximum (θ0) and the angle for the first side maximum of the longer wavelength (θ2). Mathematically, we can express this as:

θ1 = (θ0 + θ2) / 2

Now, let's use the small angle approximation to convert the angles to radians. Since tanθ ≈ sinθ for small angles, we have:

tanθ1 ≈ θ1

tanθ0 ≈ θ0

tanθ2 ≈ θ2

Substituting these approximations into the equation above, we get:

tanθ1 = (tanθ0 + tanθ2) / 2

From here, we can rearrange the equation to solve for θ0:

θ0 = 2tanθ1 - tanθ2

Finally, we can find the wavelengths λ1 and λ2 using the relationship between angle and wavelength:

θ = λ / L

Here, θ is the angle (in radians), λ is the wavelength, and L is the distance between the grating and the screen. We can rearrange the equation to solve for λ:

λ = θ * L

Substituting the values of θ0, θ1, θ2, and L into the equation, we can calculate the wavelengths λ1 and λ2.