Refraction at a boundaryRunning
Optics bench. Incidence angle at 73.6 °. Incident side 0.959 , Refracted side 0.634 . Currently incident side is over by 0.325.TUNINGn₁ = 1n₂ = 1.5needed 25°θ₁ = 73.6°θ₂ = 39.76°
t = 0.0s

Incident side

0.959

Refracted side

0.634

Incident side is over by 0.325

73.6 °

Optics bench. Incidence angle at 73.6 °. Incident side 0.959 , Refracted side 0.634 . Currently incident side is over by 0.325.

What has to balancen1sin⁡θ1=n2sin⁡θ2n_1\sin\theta_1 = n_2\sin\theta_2

What is happening

You take control of a fibre-optic splice being aligned against its acceptance cone.

The internal ray is off its mark. It must sit at 25° to the normal, and light passes from air at n = 1.00 into glass at n = 1.50.