A red laser is fired at the surface of a glass block, striking the surface at a

Question

A red laser is fired at the surface of a glass block, striking the surface at a 35 degree angle relative to the normal. Knowing the index of refraction of the glass block n= 1.5, the light ray will bend:

(Hint: Use Snell's Law to find the new, refracted angle. Is it smaller or greater than 35 degrees?)

toward the normal

away from the normal

the light ray will not bend at all

impossible to tell from the information given

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?)

40 degrees (the light ray doesn't bend)

46.5 degrees

34.7 degrees

58.4 degrees

toward the normal

away from the normal

the light ray will not bend at all

impossible to tell from the information given

Explanation / Answer

snell's law :

n1*sin(theta1)=n2*sin(theta2)

where n1 and n2 are index of refraction and theta1 and theta2 are angle of incidence and angle of refraction respectively

here n1=1

theta1=35 degrees

n2=1.5

then theta2=22.481 degrees

so it will bend towards the normal

option a is correct.

Q2.

so for water, angle of incidence=40 degrees

hence n1=1.5

theta1=40 degrees

n2=1.33

then theta2=46.465 degrees=46.5 degrees

hence second option is correct.

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