Question:
A circular disc of radius ‘R’ is placed co-axially and horizontally inside an opaque hemispherical bowl of radius ‘a’. The far edge of the disc is just visible when
viewed from the edge of the bowl. The bowl is filled with transparent liquid of refractive index µ and the near edge of the disc becomes just visible. How far below
the top of the bowl is the disc placed?
Solution:
Distance at which the bowl should be placed in the disc is given as:
d = μ(a2 – b2)/√(a + r)2 – μ(a – r)2
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.