A circular disc of radius 'R' is placed co-axially and horizontally inside an opaque hemispherical bowl of radius 'a' (Fig). 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?
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From the given figure, we have
AM is the direction of incidence ray before liquid is filled. After filling the bowl with the liduid, BM is the direction of the incidnet ray, Refracted ray in both cases is same as that along AM.
Let the disc is separated by O at a distance d as shown in figure.
From the figure, we have
N = 90º, OM = a, CB = NB = a - R, AN = a + R
Now, $\sin\text{t}=\frac{\text{a}-\text{R}}{\sqrt{\text{d}^2+(\text{a}-\text{R})^2}}\text{ and }\sin\alpha=\cos(90^\circ-\alpha)=\frac{\text{a}+\text{R}}{\sqrt{\text{d}^2+(\text{a}-\text{R})^2}}$
According to Snell's law we, have
$\frac{1}{\mu}=\frac{\sin\text{i}}{\sin\text{r}}=\frac{\sin\text{i}}{\sin\alpha}$
Substituting $\sin\text{i}=\frac{\text{a}-\text{R}}{\sqrt{\text{d}^2+(\text{a}-\text{R})^2}}\text{ and }\sin\alpha=\frac{\text{a}+\text{R}}{\sqrt{\text{d}^2+(\text{a}-\text{R})^2}}$, we get
$\text{d}=\frac{\mu(\text{a}^2-\text{R}^2)}{\sqrt{(\text{a}+\text{R})^2-\mu(\text{a}-\text{R})^2}}$
AM is the direction of incidence ray before liquid is filled. After filling the bowl with the liduid, BM is the direction of the incidnet ray, Refracted ray in both cases is same as that along AM.
Let the disc is separated by O at a distance d as shown in figure.
From the figure, we have
N = 90º, OM = a, CB = NB = a - R, AN = a + R
Now, $\sin\text{t}=\frac{\text{a}-\text{R}}{\sqrt{\text{d}^2+(\text{a}-\text{R})^2}}\text{ and }\sin\alpha=\cos(90^\circ-\alpha)=\frac{\text{a}+\text{R}}{\sqrt{\text{d}^2+(\text{a}-\text{R})^2}}$
According to Snell's law we, have
$\frac{1}{\mu}=\frac{\sin\text{i}}{\sin\text{r}}=\frac{\sin\text{i}}{\sin\alpha}$
Substituting $\sin\text{i}=\frac{\text{a}-\text{R}}{\sqrt{\text{d}^2+(\text{a}-\text{R})^2}}\text{ and }\sin\alpha=\frac{\text{a}+\text{R}}{\sqrt{\text{d}^2+(\text{a}-\text{R})^2}}$, we get
$\text{d}=\frac{\mu(\text{a}^2-\text{R}^2)}{\sqrt{(\text{a}+\text{R})^2-\mu(\text{a}-\text{R})^2}}$
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