Q.A circular disc of radius $R$ is placed horizontally and coaxially inside an opaque hemispherical bowl of radius $a$, with the disc's centre on the vertical axis of the bowl. The circular rim of the bowl lies in a horizontal plane at the top, with the two ends of a rim diameter labelled A and B and its centre O; the disc sits a vertical distance $d$ below this rim plane. An observer's eye is placed at the edge (rim) of the bowl and looks across and down into it. When the bowl is empty, the far edge of the disc is just visible along the straight line of sight from the eye. The bowl is then filled to the brim with a transparent liquid of refractive index $\mu$; refraction now bends the line of sight so that the near edge of the disc becomes just visible instead. Find how far below the top of the bowl (the value of $d$) the disc is placed, in terms of $a$, $R$ and $\mu$.

Set-up

Take the liquid surface (rim plane) as height $0$, with the eye at one rim edge, a horizontal distance $a$ from the axis. The disc lies at depth $d$, its centre on the axis, so its far edge is a horizontal distance $a+R$ from the eye and its near edge is $a-R$ from the eye, both at depth $d$.

Empty bowl — the line of sight to the far edge

The straight ray from the far edge to the eye makes an angle $r$ with the vertical (the surface normal):

This fixes the direction in which the eye looks.

Filled bowl — the near edge becomes visible

Keeping the same viewing direction, a ray now leaves the near edge, travels up through the liquid, and refracts at the surface (right at the rim, next to the eye) into that same direction. Inside the liquid this ray makes angle $i$ with the vertical:

Snell's law at the surface

Going from liquid to air, $\mu\sin i = \sin r$:

Solve for $d$

Square and cross-multiply: …

Method: Locating a "Just Visible" Point Using Snell's Law at a Rim/Edge

Use this method for problems where an observer's eye is fixed at one point (often the rim of a container), and a line of sight to a specific point on an object changes — or a different point becomes visible — once a medium (liquid) is added, refracting the ray.

Steps

Step 1: Fix the geometry and identify the two rays

Set up the two distances relevant to viewing: (a) the straight-line ray from the eye to the target point before refraction happens (i.e. through air, or the "empty" case), and (b) the ray that must exist after the medium is added, again ending at the eye but now bending at the interface. Express each ray's angle to the normal (vertical, for a horizontal liquid surface) using simple trigonometry — $\tan(\text{angle}) = \dfrac{\text{horizontal distance}}{\text{vertical distance}}$, converted to a sine via the right-triangle relation.

Step 2: Identify what stays fixed between the two cases

The key constraint linking the "before" and "after" scenarios is usually that the eye's position and viewing direction do not change — only what that fixed line of sight intersects changes because the ray now bends. So the angle in air (outside the medium) is the same angle in both descriptions; what differs is which point inside the medium connects to it.

Step 3: Apply Snell's law across the interface …