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NCERT Exemplar · Q26

Q.A cylindrical jar of height hh is filled to the brim with a transparent liquid of refractive index μ\mu. A small dot is marked at the centre of the bottom surface of the jar. An opaque circular disc is to be floated flat on the top surface of the liquid, centred directly above the dot (symmetric about the vertical centre-line). Find the minimum diameter of this disc so that, viewed from above, the dot cannot be seen from any direction.

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Light from the dot can only leave the top surface within a cone bounded by the critical angle. Rays outside that cone are totally internally reflected and never emerge, so to hide the dot it is enough for the disc to cover the circle where the escaping rays reach the surface. That circle has radius h/μ2−1h/\sqrt{\mu^2-1}, giving a minimum disc diameter of 2h/μ2−12h/\sqrt{\mu^2-1}.

Concept: escape cone and critical angle

A ray from the dot meets the top surface at an angle θ\theta from the vertical (the normal). Going from liquid to air it can emerge only if θ<θc\theta<\theta_c, where

sin⁡θc=1μ.\sin\theta_c=\frac{1}{\mu}.

Rays with θ≥θc\theta\ge\theta_c undergo total internal reflection and cannot carry the image of the dot out of the liquid — so they are harmless; the disc only needs to stop the rays that can escape.

Geometry

The dot is at the centre of the base, a vertical distance hh below the surface. A ray leaving the dot and meeting the surface at horizontal distance rr from the axis has

tan⁡θ=rh.\tan\theta=\frac{r}{h}.

The widest escaping ray corresponds to θ=θc\theta=\theta_c, reaching the surface at radius

r=htan⁡θc.r = h\tan\theta_c.

Evaluate tan⁡θc\tan\theta_c …

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