Question

A diffraction grating has \(4.00 \cdot 10^{3}\) slits/cm and has white light \((400 .-700 . \mathrm{nm})\) incident on it. What wavelength(s) will be visible at \(45.0^{\circ} ?\)

Short answer

Answer: The only visible wavelength at a 45.0-degree angle for this diffraction grating is 630 nm.

Step-by-step solution

Identify the formula

We need to use the formula for the diffraction grating, which relates the angle of diffraction (θ), order number (m), grating constant (d), and wavelength (λ) as follows: \( \sin{\theta} = m \cdot \frac{\lambda}{d} \)

Calculate grating constant d

Given that there are 4000 slits per centimeter, we can find the grating constant (d) by taking the inverse of this number. Convert the slit density to slits per meter for simplicity: \( d = \frac{1}{4.00 \cdot 10^{3}\,\text{slits/cm} \cdot 100\,\text{cm/m}} = \frac{1}{4.00 \cdot 10^{5}\,\text{slits/m}} = 2.50 \cdot 10^{-6}\,\text{m} \)

Rearrange formula to find the wavelength λ

Now we can rearrange the formula to find the wavelength (λ): \( \lambda = \frac{d \cdot \sin{\theta}}{m} \)

Substitute the values and find λ for different orders

We will substitute the values of d and θ into the equation and find the wavelength (λ) for different orders (m), keeping in mind that the visible spectrum lies between 400 nm and 700 nm: For m = 1: \( \lambda_1 = \frac{(2.50 \cdot 10^{-6}\,\text{m}) \cdot \sin{45.0^{\circ}}}{1} = 630\,\text{nm} \) For m = 2: \( \lambda_2 = \frac{(2.50 \cdot 10^{-6}\,\text{m}) \cdot \sin{45.0^{\circ}}}{2} = 315\,\text{nm} \)

Identify the visible wavelengths

Now that we have the wavelengths for the first two orders (m = 1 and m = 2), we can identify which of them fall within the visible light spectrum (400 nm to 700 nm). λ1 = 630 nm, which falls within the visible spectrum. λ2 = 315 nm, which falls outside the visible spectrum. So the only visible wavelength at a 45.0-degree angle for this diffraction grating is 630 nm.