Optics - Beam illuminating coin under water

AI Thread Summary
The discussion centers on determining the correct angle for a light beam to illuminate a coin submerged in water. The original answer provided is 28.90 degrees, but there is confusion as another participant calculates it to be 43.29 degrees. Clarification is sought on whether the light ray is reflecting off the wall or refracting through the water. One participant notes that if the angle in water is approximately 31 degrees, the corresponding angle in air is about 43.18 degrees. The conversation highlights the importance of understanding the principles of reflection and refraction in optics.
frozen7
Messages
163
Reaction score
0
A coin lies on the bottom of a pool under 1.5 m
of water and 0.90 m from the side wall as shown
in Figure 2. If a light beam is incident on the
water surface at the wall,at what angle relative
to the wall must the beam be directed so it will
illuminate the coin. Ans: 28.90

The answer I get is 43.29, different with the given answer? May I know which answer is correct actually?
Thanks.
 
Physics news on Phys.org
I don't see a diagram ... but if :
the ray's angle in the water (measured from the vertical) is about 31 degrees,
then the ray's angle in air (measured from the vertical) is about 43.18 degrees.
Are they REFLECTING the ray off the wall? (I expected this to be a refraction)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top