51. A light ray is incident on a flat piece of glass with index of refraction n as in Fig. 32-24. Show that if the incident angle 0 is small, the emerging ray is displaced a distance d = t0(n – 1)/n, where t is the thickness of the glass, 0 is in radians, and d is the perpendicular distance between the incident ray and the (dashed) line of the emerging ray (Fig. 32–24).
51. A light ray is incident on a flat piece of glass with index of refraction n as in Fig. 32-24. Show that if the incident angle 0 is small, the emerging ray is displaced a distance d = t0(n – 1)/n, where t is the thickness of the glass, 0 is in radians, and d is the perpendicular distance between the incident ray and the (dashed) line of the emerging ray (Fig. 32–24).
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P-3 Please help me with the below question with step by step explanation, please.
NOTE: note that d is the perpendicular distance between the incident ray (pink solid line) and the pink dashed line
Transcribed Image Text:51. (III) A light ray is incident on a flat piece of glass with
index of refraction n as in Fig. 32-24. Show that if the
incident angle 6 is small, the emerging ray is displaced a
distance d = t0(n – 1)/n, where t is the thickness of the
glass, 0 is in radians, and d is the perpendicular distance
between the incident ray and the (dashed) line of the
emerging ray (Fig. 32-24).
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