Question

A diffraction grating, ruled with 300 lines per mm, is illuminated with a white light source at normal incidence.

Answer 1

the expression for diffraction grating allows to find the results for the questions for the angular separation are:

i) The third order is Δθ = 0.203 rad.

ii) The first order with water is Δθ = 0.046 rad.

The diffraction grating is a system formed by a large number of equally spaced lines whose diffraction is given by the expression.

          d sin θ = m λ

Where d is the distance between two lines, θ is the angle of diffraction, the order of diffraction and λ is the wavelength.

i) Let's start by looking for the separation between two lines

Let's use a rule of direct proportions. If there are 300 lines in 1 mm, what distance is there between two lines.

         d = 1 lines (1 mm / 300 lines) = 3,333 10⁻³ mm

         d = 3.333 10⁻⁶ m

Let's find the angle of diffraction for the third order (m = 3) for each wavelength.

λ₁ = 400 nm = 400 10⁻⁹ m

         sin θ₁ = $\frac{m \ \lambda }{d}$m λ/ d

         sin θ₁ = $\frac{3 \ 400 \ 10^{-9} }{3.333 \ 10^{-6} }$  

         θ₁ = sin⁻¹ 0.3600

         θ₁ = 0.368 rad

λ₂ = 600 nm = 600 10⁻⁹ m

         sin θ₂ = $\frac{3 \ 600 \ 10^{-9} }{3.333 \ 10^{-6} }$  

         θ₂ = sin⁻¹ 0.5401

         θ₂ = 0.571 rad

The angular separation is

         Δθ = θ₂ - θ₁

         Δθ = 0.571 - 0.368

         Δθ = 0.203 rad

ii) In this case, the separation between the network and the observation screen is filled with water.

When the rays leave the network they undergo a refraction process, for which they must comply with the relationship.

           $n_i \ sin \theta_1 = n_r \ sin \theta_r$

The incident side is in the air, therefore its refractive index is n_i = 1 and when it passes into the water with refractive index n_r = 1.33.

Let's start looking for the incident angles for the first order of diffraction.

      m = 1

λ₁ = 400 nm

         θ₁ = sin⁻¹  $\frac{1 \ 400 \ 10^{-9}}{3.33 \ 10^{-6}}$

         θ₁ = 0.120 rad

λ₂ = 600 nm

        θ₂ = sin⁻¹¹ $\frac{1 \ 600 \ 10^{-9} }{3.33 \ 10^{-6}}$

        θ₂ = 0.181 rad

we use the equation of refraction.

         $\theta_r$  = sin⁻¹ ($\frac{n_i}{n_r} \ sin \ \theta_i$ )

λ₁ = 400 nm  

       θ₁ = sin¹ ($\frac{1 sin 0.120}{1.33}$

       θ₁ = 0.090 rad

λ₂ = 600 nm

        θ₂ =sin⁻¹  $\frac{1 sin 0.181}{1.33}$

        θ₂ = 0.1358 rad

The angular separation is

          Δθ = 0.1358 - 0.090

          Δθ = 0.046 rad.

In conclusion using the relation for the diffraction grating we can find the results for the questions about angular separation are:

       i) The third order is Δθ = 0.203 rad.

      ii) The first order with water is Δθ = 0.046 rad.

Learn more here: brainly.com/question/473160