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  • A circular disc of radius ‘R’ is placed co-axially and horizontally inside an opaque hemispherical bowl of radius ‘a’ (Fig. 9.5). The far edge of the disc is just visible when viewed from the edge of the bowl.

A circular disc of radius ‘R’ is placed co-axially and horizontally inside an opaque hemispherical bowl of radius ‘a’ (Fig. 9.5). 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 \mu and the near edge of the disc becomes just visible. How far below the top of the bowl is the disc placed?

 

 

Answers (1)

As shown in the figure, AM is in the direction of the incidence ray before filling the liquid. After filling liquid in the bowl, BM is in the direction of the incident ray. Refracted in both cases is same as that along MN.

Let the disc is separated by O at a distance d as shown in the figure. Also, considering the angle

P=90^{o}, OM=a,CB=R,BP=a{-}R,AP=a+R

Here, in  \Delta BMP,

\sin\; i=\frac{BP}{BM}=\frac{a-R}{\sqrt{d^{2}+\left ( a-R \right )^{2}}}\; \; \; \; \; \; \; \; \; \; \; ....(i)

and in

\Delta AMP\; \cos \left ( 90^{o}-\alpha \right )=\sin \; \alpha =\frac{a+R}{\sqrt{d^{2}+\left ( a+R \right )^{2}}}\; \; \; \; \; \; \; \; \; \; \; \; .....(ii)

But on applying Snell's law at point M

\mu \times \sin \; i=1 \times \; sin \; r

\mu \times \frac{a-R}{\sqrt{d^{2}+\left ( a-R \right )^{2}}}=1\times \frac{a+R}{\sqrt{d^{2}+\left ( a+R \right )^{2}}}

\Rightarrow d=\frac{\mu \left ( a^{2}-b^{2} \right )}{\sqrt{\left ( a+r \right )^{2}-\mu \left ( a-r \right )^{2}}}

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infoexpert23

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