Determine the displacement of a collimated beam incident on a tilted plane-parallel plate, or window. This displacement is dependent on the angle of incidence, the thickness of the plate, and the refractive index of the plate. This calculator also determines the displacement of the secondary reflection resulting from a small portion of the beam reflecting off the glass-air interface of the plate.

The exit beam will be parallel to the input beam and magnification remains constant when imaging through a plane-parallel plate.

Beam Displacement, s (mm) : **-- **

Displacement of Secondary Reflection, x (mm): **-- **

**Note:** This calculator is valid within the following limits:

• t > 0mm

• 0° < θ < 90°

• n > 0

• 0° < θ < 90°

• n > 0

$$ s = t \cdot \sin{\theta} \left( 1 - \frac{\cos{\theta}}{\sqrt{n^2 - \sin^2{\theta}}} \right) $$ |

$$ x = \frac{t \cdot \sin{\left( 2 \theta \right)}}{\sqrt{n^2 - \sin^2{\theta}}} $$ |

s |
Beam Displacement |

x |
Displacement of secondary reflection |

θ |
Angle of incidence between beam and plate normal |

t |
Thickness of plate |

n |
Refractive index of medium |

**Question:** What is the displacement of a beam and its secondary reflection when the beam passes through a 10mm thick N-BK7 plate at an angle of incidence of 15 degrees?

**Answer:** The beam displacement and secondary reflection displacement can be determined by:

$$ s = \left( 10 \text{mm} \right) \sin{\left( 15° \right)} \left( 1 - \frac{\cos{\left( 15° \right)}}{\sqrt{1.517^2 - \sin^2{\left(15° \right)}}} \right) = 0.916 \text{mm} $$

$$ x = \frac{\left( 10 \text{mm} \right) \sin{\left( 2 \cdot 15° \right)}}{\sqrt{1.517^2 - \sin^2{\left( 15° \right)}}} = 3.499 \text{mm} $$

The displacement of the beam is 0.916mm and the displacement of the beam’s secondary reflection is 3.499mm.