How does the presence of water affect the image formed by a converging lens?

[apunisheriii]

Homework Statement

A needle of length 5.0cm placed at the bottom of a beaker and a converging lens of focal length 20.0cm is held 30.0cm from the needle.The needle is illuminated with a lamp from the bottom.An image of the needle is formed on a screen.The beaker is now filled with water(refractive index=4/3)to a depth of 20cm.What change can be observed in the image?
If the screen is moved to get a new sharp image,what will the image distance and object be in this new position??

i squeezed my brain so hard..still don hv any idea at all..becoz there's a space (air)between the lens and the water.When i am using the lens maker's formula,im confusing whether i should assume there r three lens -water,converging lens and water or two or one..
can somebody give some hints about how to solve it??

 

[Doc Al]

Forget about the lens for a moment. From the viewpoint of someone looking down into the beaker (from the position of the lens), will the addition of the water make the apparent position of the needle closer or farther away? By how much?

 

[apunisheriii]

oh!
the apparent depth is 22.5cm from the lens
and let u=22.5cm and solve it by lens formula
is it correct??

 

[apunisheriii]

sorry...
mistake...real depth shouldn't include the air...u=25cm
by the way!thnx!

 

[Doc Al]

the apparent depth is 22.5cm from the lens

How did you get that value? Realize that the water is only 20 cm deep.

 

[Doc Al]

sorry...
mistake...real depth shouldn't include the air...u=25cm
by the way!thnx!

Now you've got it.