The glass block (n = 1.50) is placed on the surface of a container of water (n =

Question

The glass block (n = 1.50) is placed on the surface of a container of water (n = 1.33). The beam of the red laser is adjusted so that it reaches the other side of the block at a 40 degree angle relative to the normal before refracting again as it goes from glass to water. At what angle relative to the normal does the light beam travel through the water? (Hint: Now, do Snell's Law again but the ray goes from glass to water. Use the new value for the incident angle. What's the refracted angle in this case?)

Explanation / Answer

here,

for rectangular block

angle of refraction will be same as angle of incidence in water

angle of incidence , i = 40 degree

n1 = 1.50

n2 = 1.33

Using Snell's law

n1 * sin(i) = n2 * sin(r)

1.50 * sin(40) = 1.33 * sin(r)

solving for r

r = 46.5 degree

the angle of refraction in water is 46.5 degree

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