Question

We take a rectangular glass slab with a thickness of 13 mm. There are perfectly aligned rulers on the two long surfaces, ruler A and ruler B. The glass has an index of refraction of 1.3 with no dispersion. We aim a laser at the zero mark of ruler A at an angle of 60° to a normal line that is perpendicular to ruler A. It refracts through the glass and exit from the other side, through ruler B. At what distance (in mm) from the zero mark of ruler B does the laser beam exit?

Answer #1

By law of refraction , we have , [ sin (i) / sin(r) ] =

where i = 60^{o} is angle of incidence and r is angle of
refraction and
= 1.3 is refractive index of glass.

hence angle of refraction , sin(r) = sin(60) / 1.3 = 0.666

Hence , angle of refraction r = sin^{-1} (0.666)
42^{o}

As seen from figure , tan(42) = x /13 or x = tan42 13 = 11.61 mm

Hence laser beam exit around 12 mm from zero mark

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