Difference in depth between the swimmer and its image

AI Thread Summary
The discussion focuses on calculating the apparent difference in depth between a swimmer and their image in a plane mirror located at the bottom of a 3 m deep swimming pool. The formula used is h' = h(1/(4/3)), resulting in an apparent depth of 0.75 m for the swimmer and 2.25 m for the mirror. The difference calculated between these two depths is 1.5 m, but it is noted that the total actual distance is 4 m, leading to an apparent depth of 3 m. The final answer of 3 m is confirmed as correct, with a suggestion to simplify the calculation process. The discussion emphasizes the importance of understanding the relationship between actual and apparent depths in optics.
jsalapide
Messages
39
Reaction score
0
A plane mirror is placed on the level bottom of a swimming pool, which holds water to a depth of 3 m. A swimmer is located 2.0 m above the mirror. An observer above the water looks vertically downward at the swimmer and its image in the mirror. What is the apparent difference in depth between the swimmer and its image in the mirror?

I used the formula h'=h(1/(4/3)) where h' is the apparent depth as seen by the observer and h as the actual depth of the object.

the h' of the swimmer is 0.75 m while the mirror is 2.25 m.

I subtracted the 0.75 to 2.25 to get the distance between those 2. And I got 1.5.

Since it's a plane mirror, the image distance is only twice the object distance from the mirror.

My answer is 3 m..

is it correct?
 
Physics news on Phys.org
Hi jsalapide! :smile:
jsalapide said:
… I used the formula h'=h(1/(4/3)) where h' is the apparent depth as seen by the observer and h as the actual depth of the object.

the h' of the swimmer is 0.75 m while the mirror is 2.25 m.

I subtracted the 0.75 to 2.25 to get the distance between those 2. And I got 1.5.

Since it's a plane mirror, the image distance is only twice the object distance from the mirror.

My answer is 3 m..

is it correct?

Yup! :smile:

(but a bit longwinded … you could have said the actual distance is 4m, then apply the h' formula, which reduces it to 3m … you don't need to find the 0.75 :wink:)
 
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