Radius of curvature of glass and water lens

[desmond iking]

Homework Statement

by taking the lower curvature as r1 , and the upper curvature as r2 ,
i don't know whether r1 is 20cm , r2 is 10cm or vice versa.

But according to the ans r1= 10 cm , r2= 20cm . why is it so?

Homework Equations

The Attempt at a Solution

 

[haruspex]

Draw two circles the same size, with centres only a little apart. look at where the circles meet in relation to the two centres. The line joining the the intersections bisects the line joining the centres.
If the circles have different radii then the intersections will be closer to one of the circle centres - which one?

 

[desmond iking]

Draw two circles the same size, with centres only a little apart. look at where the circles meet in relation to the two centres. The line joining the the intersections bisects the line joining the centres.
If the circles have different radii then the intersections will be closer to one of the circle centres - which one?

i drew the diagram , and it show what you have said. but i still don't understand why it is so . can you explain further?

 

[haruspex]

i drew the diagram , and it show what you have said. but i still don't understand why it is so . can you explain further?

Let me try another approach.
Wrt the original diagram, the question is whether r1 can be the larger radius.
Start with two circles of different radii, with the smaller inside and touching the larger. For reference, take the circle centres to be on the same X axis, with the point of contact on the left.
If we shift the smaller to the left it will be partly outside, and the left half of the arrangement will look like the original diagram but with r1 as the smaller radius. So see what happens if we shift it to the right.
At first, the smaller will be wholly inside the larger, no intersection. When intersection occurs again, the right half will again look like the original diagram, but again, with r1 as the smaller radius.
If we continue moving the smaller to the right, until it is nearly outside the larger, neither the left half nor the right half of the arrangement looks quite like the original diagram. In each case, the lens stretches more than half way around the circles.

You may think that's a lot to have to go through in order to solve the given problem. I didn't have to go through all that to solove it, because I just looked at the diagram and saw that r1 was the tighter curve, so must have the smaller radius. But I couldn't see a way to turn that into a solid argument.

 

[HalfLostAndSearching]

I would first like to clarify that the radius of curvature for a lens is measured from the center of the lens to the surface of the lens. This is different from the radius of curvature for a spherical object, which is measured from the center of the sphere to its surface.

Now, in the case of a glass and water lens, the radius of curvature will depend on the refractive indices of both materials. The refractive index of glass is higher than that of water, which means that light will bend more when passing through the glass lens. This results in a smaller radius of curvature for the lower surface of the lens (r1) compared to the upper surface (r2).

To determine the exact values of r1 and r2, we can use the lens maker's formula, which relates the radius of curvature, refractive index, and focal length of a lens. This formula is given by:

1/f = (n2 - n1) * (1/r1 - 1/r2)

Where f is the focal length, n1 and n2 are the refractive indices of the two materials, and r1 and r2 are the radii of curvature for the lower and upper surfaces, respectively.

In this case, we know that the focal length of the lens is positive (since it is a converging lens), and the refractive index of glass is higher than that of water. Therefore, to get a positive focal length, we need the term (1/r1 - 1/r2) to be positive. This means that r1 must be smaller than r2, which is why the answer given is r1=10 cm and r2=20 cm.

In conclusion, the values of r1 and r2 depend on the refractive indices of the materials and can be determined using the lens maker's formula. The answer given is correct and follows the laws of optics.