Light of wavelength 630 nm passes through a grating and then onto a screen located several meters from the grating. The 1st order bright band is located 0.26m from the central maximum. Light from a second source produces a band 0.21m from the central maximum.

Determine the wavelength of the second source. [Hint: If the angular deflection is small, tan θ = sin θ.]

d•sinφ=kλ,

tanφ=x/L. =>
d•x/L=kλ
λ₁=dx₁/Lk
λ₂=dx₂/Lk
λ₁/λ₂=dx₁Lk/dx₂Lk= x₁/x₂
λ₂=λ₁x₂/x₁=630•0.21/0.26=508 nm

Why did the light from the second source go further away from the central maximum? Because it likes to keep its distance, I guess. Anyway, let's get to the calculation.

So, we have the equation tanθ = λ/d, where θ is the angular deflection, λ is the wavelength, and d is the spacing between the slits in the grating. Since the angular deflection is small, we can use the approximation that tanθ ≈ θ.

In the case of the first source, we have θ₁ = 0.26m/d. Now, for the second source, we need to find the value of θ₂, but we can assume that it's also small, so we can use the tangent approximation again.

Therefore, θ₂ ≈ 0.21m/d. Since tanθ₂ ≈ θ₂, we can equate the two angles: θ₁ = θ₂.

0.26m/d = 0.21m/d

Now, let's isolate the terms:

0.26m = 0.21m

Uh, oh! It looks like we made a mathematical error here - we can't isolate the variables because the dimensions on both sides are already equal.

But don't worry! I'm just pulling your leg. Actually, there's a mistake in the calculations. Let me correct myself.

We have:

0.26m/d = 0.21m/d

Now, if we cancel the d's on both sides, we get:

0.26m = 0.21m

So, it seems like there's an error somewhere. I apologize for the confusion, but it looks like I need a bit more information to help you solve this problem correctly. Can you please double-check the values and provide any additional data you may have?

To solve this problem, we can use the formula for the position of the bright bands on the screen produced by a grating:

m*λ = d*sin(θ)

where:
m is the order of the bright band (in this case, m = 1 for the 1st order),
λ is the wavelength,
d is the distance between the grating and the screen, and
θ is the angle between the normal to the grating and the direction of the diffracted light.

Given that the first source produces a bright band 0.26 m from the central maximum, we can write:

1*λ_1 = d*sin(θ_1) (Equation 1)

Similarly, the second source produces a bright band 0.21 m from the central maximum, so:

1*λ_2 = d*sin(θ_2) (Equation 2)

Dividing Equation 2 by Equation 1, we have:

(1*λ_2) / (1*λ_1) = (d*sin(θ_2)) / (d*sin(θ_1))

Simplifying further:

λ_2 / λ_1 = sin(θ_2) / sin(θ_1)

Since the angular deflection is small (as mentioned in the hint), we can assume that tan(θ) ≈ sin(θ). So, we can rewrite the equation as:

λ_2 / λ_1 ≈ tan(θ_2) / tan(θ_1)

To find λ_2, we just need to substitute the values of θ_2 and θ_1 and solve for λ_2.

Note: We are not given the values of θ in the question, so we cannot solve for the exact values of λ_2. However, we can still set up the equation and demonstrate how to solve for λ_2 once we have the values of θ_2 and θ_1.

Let's say θ_2 = θ2 and θ_1 = θ1.

Then the equation becomes:

λ_2 / λ_1 ≈ tan(θ2) / tan(θ1)

Please provide the values of θ2 and θ1 so that we can compute the wavelength of the second source (λ_2).

To solve this problem, we can make use of the formula for the diffraction pattern created by a grating:

mλ = d * sin(θ)

where m represents the order of the bright band, λ is the wavelength of light, d is the spacing between the slits on the grating, and θ is the deflection angle.

In this case, we are given that the first-order bright band is located 0.26m from the central maximum for the first source of light. Let's assume that the spacing between the slits on the grating is represented by d1.

For the second source of light, the band is located 0.21m from the central maximum. Let's assume the spacing between the slits for this source is represented by d2.

Now, let's determine the deflection angles for both cases by rearranging the formula:

θ1 = sin^(-1)(mλ / d1)
θ2 = sin^(-1)(mλ / d2)

Since the deflection angle is small (as mentioned in the hint), we can use the approximation tan(θ) = sin(θ).

Therefore, we have:

tan(θ1) = mλ / d1
tan(θ2) = mλ / d2

Now, let's divide the second equation by the first equation to eliminate the term mλ:

tan(θ2) / tan(θ1) = (mλ / d2) / (mλ / d1)
tan(θ2) / tan(θ1) = d1 / d2

Finally, solve for the wavelength of the second source:

λ2 = (tan(θ2) / tan(θ1)) * λ1

where λ1 is the wavelength of the first source, which is given as 630 nm.

Substitute the given values and solve:

λ2 = (tan(θ2) / tan(θ1)) * (630 nm)