A light ray is incident on a flat piece of glass with index of refraction n as in Fig 23 22 Show that if the incident angle is small, the emerging ray is displaced a distance d t (n 1) n from the incident ray, where t is the thickness of the glass and is in radians ...

Because the glass surfaces are parallel the exit beam will be traveling in the same dire ...

A light ray is incident on a flat piece of glass with index of refraction n as

Question:

A light ray is incident on a flat piece of glass with index of refraction n as in Fig. 23-22. Show that if the incident angle θ is small, the emerging ray is displaced a distance d = t θ (n – 1) / n from the incident ray, where t is the thickness of the glass and θ is in radians.