Refraction Angle: Find the Answer to Your Question

  • Thread starter Thread starter StudentofPhysics
  • Start date Start date
  • Tags Tags
    Angle
AI Thread Summary
The discussion centers on calculating the angle of refraction for a ray of light passing from a glass block into liquid carbon disulfide. The initial angle of incidence is 46.0°, and the refractive indices are provided (glass n = 1.52, liquid n = 1.63). The incorrect initial calculation of 50.48° was clarified to be the angle within the glass, not the angle of emergence. To find the correct angle as the ray exits into the liquid, the angle must be adjusted by subtracting from 90°, leading to an angle of 39.52° before applying Snell's law again. The final understanding emphasizes the need to correctly apply the indices and angles at each interface.
StudentofPhysics
Messages
65
Reaction score
0
The drawing shows a rectangular block of glass (n = 1.52) surrounded by liquid carbon disulfide (n = 1.63). A ray of light is incident on the glass at point A with a = 46.0° angle of incidence. At what angle of refraction does the ray leave the glass at point B?

p26-20alt.gif



I know to use n1 sin 01 = n2 sin 02 (0 = theta)

i came up with 50.48 degrees, which was wrong. Then I figured since it was traveling back out of the block, maybe i should keep going and solve for that angle, which just cam out to be 46 degrees, also wrong.

What am I doing wrong?
 
Physics news on Phys.org
StudentofPhysics said:
i came up with 50.48 degrees, which was wrong. Then I figured since it was traveling back out of the block, maybe i should keep going and solve for that angle, which just cam out to be 46 degrees, also wrong.

50.48° is the angle at which the ray emerges in the block. It then continues its course and hits the surface block-liquid at the angle 90° - 50.48° = 39.52° (draw a little square-triangle to see it). Then you got to run another n1 sin 01 = n2 sin 02 to find the angle 02 at which the ray emerges in the liquid. But this time, of course, the index "1" refers to the glass and the index "2" refers to the liquid.
 
Thank you. I forget to subtract 50.48 from 90 before continuing on.
Got it now though.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .

Similar threads

Back
Top