Show that for a material with refractive index µ ≥ 2 , light incident at any angle shall be guided along a length perpendicular to the incident face.
Consider a rectangular slab of refractive index μ≥2. An incidence ray incidence at angle i on face PQ at incidence point. Refracted ray BC strike at face QR which is perpendicular to PQ with incidence angle ic so that refracted ray CD passes normal to the face PQ as per required in question. So ic must be critical angle. μ=sinic1 (Snell's law at c) sinic≥μ1[sinic=μ1] sin(90−r)≥μ1[∵r+90+ic=180] cosr≥μ1 cos2r≥μ21 (squaring both sides) −cos2r≤−μ21 1−cos2r≤1−μ21 sin2r≤cos2r≤1−μ21...(I) sini=μ or sini=μsinr⇒sin2i=μ2sin2r (on squaring on both sides) μ21sin2i=sin2r...(II) Put (II) in (I) μ21sin2i≤1−μ21 sin2i≤μ2−1 (on multiplying by μ2 on both sides) For smallest angle ie., i=90 sin290≤μ2−1[∵μ≥2] 1+1≤μ2 2≤μ2 Taking square root 2≤μ Hence proved
