The Spinning Beaker
A Fluid Dynamics Primer
Imagine a beaker filled with water, spinning gracefully on a horizontal table. The water doesn't stay flat; the centrifugal pseudo-force pushes the fluid radially outward, causing it to climb up the walls. This creates a beautiful curved surface known as a paraboloid.
Let's focus on the dimensions given in the problem. The depth of the water at the exact center is H, and the water rises by an extra height h at the edges. The radius of the beaker is r. The problem asks us to treat this curved surface near the center as a part of a sphere with a radius of curvature R.
Unlocking the Geometry of the Surface
If we look at the geometry of this spherical cap, we can use the Pythagorean theorem on the right-angled triangle formed by the radius R, the horizontal distance r, and the vertical drop (R−h).
Let's expand that equation.
The R2 terms cancel out beautifully. Rearranging this gives us the radius of curvature R:
This perfectly matches our first option!
Now, let's bring in some fluid statics. The shape of a rotating liquid surface is governed by the equation y=2gω2x2. At the edge, where x=r, the height is h. So, h=2gω2r2. Since the problem states that h≪r, the h2 term in our radius equation becomes negligible. Our radius R simplifies to 2hr2, which is exactly ω2g.
The Optics of a Liquid Paraboloid
Next, we need to find the apparent depth of the bottom of the beaker. We'll use the master equation for refraction at a single spherical surface. Light travels from the water (our medium 1, with μ1=34) into the air (our medium 2, with μ2=1).
Sign conventions are critical here! Don't make a silly mistake. The object is at the bottom of the beaker, so u=−H. The light travels upwards, and the center of curvature of this "cupped" surface is also above the surface in the air. Therefore, R is positive. Let's substitute these values into our formula.
The Final Calculation
Apparent Depth
Let's simplify the terms. The double negative on the left becomes a positive.
Moving it to the right side, we can factor out −3H4.
v1=−3H4−3R1=−3H4(1+4RH)
Remember that value of R we found earlier? Let's bring that back. Substituting R=ω2g into our equation, the term 4RH transforms into 4gω2H.
Finally, we just invert both sides to find the image position v.
The negative sign simply means the image is virtual and formed below the surface, which is exactly what we expect for apparent depth. The magnitude perfectly matches option (D). What an elegant result combining fluid mechanics and optics!
