Question

A 610 lines/mm diffraction grating is in an empty aquarium tank. The index of refraction of the glass walls is nglass = 1.50. A helium-neon laser (λ = 633nm) is outside the aquarium. The laser beam passes through the glass wall and illuminates the diffraction grating. What is the first-order diffraction angle of the laser beam? What is the first-order diffraction angle of the laser beam after the aquarium is filled with water ( nwater = 1.33)?

Answer 1

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Answer 2

First-Order Diffraction Angle

Given:

• Diffraction grating density: $610$ lines/mm → Grating spacing $d=\frac{1}{610\times 10^3}\text{m}=1.639\times 10^{-6}\text{m}$.

• Laser wavelength (in air): ${\lambda }_{\text{air}}=633\text{nm}=633\times 10^{-9}\text{m}$.

• Refractive indices:

• Glass ($n_{\text{glass}}$): 1.50

• Water ($n_{\text{water}}$): 1.33

• 
  

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Step 1: Empty Aquarium (Glass Walls Only)

1. Wavelength in glass:

   $${\lambda }_{\text{glass}}=\frac{{\lambda }_{\text{air}}}{n_{\text{glass}}}=\frac{633\text{nm}}{1.50}=422\text{nm}\mathrm{.}$$

2. Diffraction condition (first order, $m=1$):

   $$d\sin \theta =m\lambda ⟹\sin \theta =\frac{{\lambda }_{\text{glass}}}{d}\mathrm{.}$$$$\sin \theta =\frac{422\times 10^{-9}}{1.639\times 10^{-6}}\approx \mathrm{0.2575.}$$

3. First-order angle:

   $$\theta ={\sin }^{-1}(0.2575)\approx 14.92^{\circ }\boxed{{\displaystyle 14.9^{\circ }}}\mathrm{.}$$

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Step 2: Aquarium Filled with Water

1. Wavelength in water:

   $${\lambda }_{\text{water}}=\frac{{\lambda }_{\text{air}}}{n_{\text{water}}}=\frac{633\text{nm}}{1.33}\approx 476\text{nm}\mathrm{.}$$

2. Diffraction condition:

   $$\sin {\theta }^{\mathrm{\prime }}=\frac{{\lambda }_{\text{water}}}{d}=\frac{476\times 10^{-9}}{1.639\times 10^{-6}}\approx \mathrm{0.2904.}$$

3. First-order angle:

   $${\theta }^{\mathrm{\prime }}={\sin }^{-1}(0.2904)\approx 16.88^{\circ }\boxed{{\displaystyle 16.9^{\circ }}}\mathrm{.}$$