Question

A light ray is passing through water (n=1.33) towards the boundary with a transparent solid at an angle of 56.4°. The light refracts into the solid at an angle of refraction of 42.1°. Determine the index of refraction of the unknown solid. How would you work this problem?

Answer 1

4 me,d ansa is 0.84.....usin d n=sin i/sin r......1.33=sin 56.4/sin r...solvin dis,r of water=38.77......dt r is d angle of refraction of light into d block is also d angle of incidence of d block..bt e=90-r.therefore,e=90-42.1=47.9again usin n=sin i/sin r.....n=sin 42.1/sin 47.9,....n=0.84