Difference in depth between the swimmer and its image

[jsalapide]

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?

 

[tiny-tim]

Hi jsalapide!

… 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!

(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 )

 

[tomcue]

Your calculations are correct, but your answer is not entirely accurate. The apparent difference in depth between the swimmer and its image in the mirror is actually 1.5 meters, not 3 meters. This means that the swimmer appears to be 1.5 meters deeper in the mirror than they actually are in the water. This is because the light rays reflecting off the swimmer's image in the mirror are coming from a virtual image that is 1.5 meters behind the mirror's surface. This creates the illusion that the swimmer is deeper in the water when viewed from above.