Calculating Apparent Depth of Print Beneath Flint Glass Plate

AI Thread Summary
The discussion revolves around calculating the apparent depth of print beneath a 3.5 cm thick flint glass plate, using the refractive index of flint glass, which is 1.66. The initial calculation suggests that the apparent depth is 2.1 cm, derived from the formula apparent depth = actual depth / refractive index. Participants emphasize the importance of confidence in optics solutions and recommend drawing a ray diagram to visualize the refraction process. They discuss Snell's law, noting that angles in air can be approximated as small, aiding in understanding the light's behavior at the glass-air interface. The conversation concludes with a reinforcement of the explanation provided.
Aoiumi
Messages
23
Reaction score
0

Homework Statement


A flint glass plate 3.5 cm thick is placed over a newspaper. How far beneath the top surface of the plate would the print appear to be if you were looking almost vertically downward through the plate?


Homework Equations


N=1.66 for flint glass
Apparent = d/n


The Attempt at a Solution


Apparent depth = 3.5 cm /1.66
= 2.1 cm

I'm not sure if this answer or the formula I used are correct. Thank you.
 
Physics news on Phys.org
You're right.
You can have more confidence in your solutions, especially in optics, if you draw the ray diagram (using Snell)
One ray travels straight up through the glass "n" into air, undeflected at the surface (sin theta=0).
The other ray goes thru the glass at angle theta_glass from the normal, but deflects as it leaves the glass,
to angle theta_air ... if these angles are small, sin theta ≈ theta.
Using Snell, that means the angle in air is ≈ theta_glass / n.
 
  • Like
Likes 1 person
Thank you for your explanation!
 
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