Optics problem with a glass lens immersed in water

[tellmesomething]

Homework Statement

attached

Relevant Equations

none

In this solution, aren't we assuming the power or inverse of focal length of these 3 lens when they are kept with respect to air??
Is that obvious in the question?

 

[TSny]

Homework Statement: attached
Relevant Equations: none

View attachment 358386

In this solution, aren't we assuming the power or inverse of focal length of these 3 lens when they are kept with respect to air??
Is that obvious in the question?

Imagine a very thin layer of air between the upper surface of the glass and the water above it and another very thin layer of air between the bottom surface of the glass and the water below it. Then all surfaces would be in contact with air.

Can you see why these imaginary thin layers of air do not significantly affect the optics of the system?

 

[tellmesomething]

Imagine a very thin layer of air between the upper surface of the glass and the water above it and another very thin layer of air between the bottom surface of the glass and the water below it. Then all surfaces would be in contact with air.

Can you see why these imaginary thin layers of air do not significantly affect the optics of the system?

I see. Focal length of Lens in another medium like water would increase, maybe not significantly, is this what you meant?

 

[TSny]

I see. Focal length of Lens in another medium like water would increase, maybe not significantly, is this what you meant?

I hope I'm interpreting your original question correctly. I believe you want to know why the solution calculates the power of each component of the compound system (water-glass-water) as though there is air on each side of the component.

Here is one way to think about it. It might not be satisfactory. Suppose we modify the system by adding very thin gaps of air between each component.

Now each component has air on both sides and the power of each component would be calculated as in the solution. But adding these gaps of air shouldn't affect how a ray of light passes through the system. So, the solution with the air gaps should be valid for the original system without the gaps.

As a step towards justifying this, consider a ray that passes from glass into water, for example.

Add a thin layer of air between the glass and the water:

Use Snell's law to show that if $\theta_1$ is the same in the two pictures, then $\theta_2$ will be the same. So, the air gap doesn't change the direction of the ray in the water. However, the air gap does displace the ray horizontally a bit:

The dotted gray ray shows the ray in the water if there is no air gap. However, you can show that the amount of sideways shift of the ray goes to zero as the thickness of the gap goes to zero. So, a very thin air gap doesn't have a significant effect on the path of the ray.

You can think about whether or not the argument still holds for curved surfaces, as in your problem.