Question

An opaque sphere of radius a is just immersed in a transparent liquid as shown in the figure. A point source is placed on a vertical diameter of the sphere at a distance $\frac{a}{2}$ from the top of the sphere. One ray originating from the point source after refraction from the air liquid interface forms tangent to the sphere. The angle of refraction for that particular ray is $30^\circ$. The refractive index of the liquid is

• A. $\frac{2}{\sqrt{3}}$
• B. $\frac{3}{\sqrt{5}}$
• C. $\frac{4}{\sqrt{5}}$
• D. $\frac{4}{\sqrt{7}}$

Answer 1

Answer: (D)

$\frac{4}{\sqrt{7}}$

Answer 2

We set up the problem by choosing a coordinate system so that the horizontal liquid–air interface is at $y=0$. “Just immersed” means that the sphere touches the interface at its top. Choosing

$$\text{Top } T=(0,0),\quad \text{Center } C=(0,a),\quad \text{Radius } = a,$$

the sphere is described by

$$x^2+(y-a)^2=a^2.$$

The point source is said to lie on a vertical line through the sphere, at a distance $\frac{a}{2}$ from the top. (It is more natural here to assume that the source is above the interface so that its light enters the liquid.) Thus, with the positive $y$–axis directed downward, we take the source as

$$S=(0,-a/2).$$

A ray from $S$ meets the interface at $R=(x_R,0)$. In air (with refractive index 1), the ray makes an incident angle

$$\sin i=\frac{|x_R|}{\sqrt{x_R^2+(a/2)^2}}.$$

After refraction at $R$ into the liquid (of refractive index $n$) the ray makes an angle $r=30^\circ$ with the vertical (normal to the interface). Snell’s law gives

$$\sin i = n\,\sin 30^\circ = \frac{n}{2}.$$

Thus,

$$\frac{|x_R|}{\sqrt{x_R^2+(a/2)^2}}=\frac{n}{2}.$$

Squaring both sides we obtain

$$\frac{x_R^2}{x_R^2+a^2/4}=\frac{n^2}{4} \quad \Longrightarrow \quad (4-n^2)x_R^2=\frac{n^2a^2}{4}.$$

Therefore,

$$x_R^2=\frac{n^2a^2}{4(4-n^2)}. \tag{A}$$

Now, the ray (in the liquid) emerging from $R$ has direction making $30^\circ$ with the vertical. Choosing (without loss of generality) the positive $x$–direction we have the direction vector

$$\hat{d}=(\sin 30^\circ,\cos 30^\circ)=(0.5,\,\sqrt{3}/2).$$

Thus the ray in the liquid is given by

$$\vec{r}(t)=(x_R+0.5\,t,\;0+(\sqrt{3}/2)\,t).$$

Let the ray touch the sphere at the tangent point $Q$. At tangency, the radius drawn to $Q$ is perpendicular to the ray’s direction. With $Q=(x_R+0.5\,t_0,\,(\sqrt{3}/2)\,t_0)$ and $C=(0,a)$, the tangency condition is:

$$\big[(x_R+0.5\,t_0)-0,\;(\sqrt{3}/2\,t_0)-a\big]\cdot (0.5,\,\sqrt{3}/2)=0.$$

Calculating the dot product:

$$0.5\,(x_R+0.5t_0)+\frac{\sqrt{3}}{2}\Big(\frac{\sqrt{3}\,t_0}{2}-a\Big)=0.$$

Note that $\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2}=\frac{3}{4}$. Hence,

$$0.5\,x_R+0.25\,t_0+0.75\,t_0-\frac{\sqrt{3}}{2}\,a=0 \quad \Longrightarrow \quad 0.5\,x_R+t_0-\frac{\sqrt{3}}{2}\,a=0,$$

so that

$$t_0=\frac{\sqrt{3}}{2}\,a-0.5\,x_R. \tag{1}$$

Since $Q$ lies on the sphere, it must satisfy

$$\Big[x_R+0.5\,t_0\Big]^2+\Big[\frac{\sqrt{3}}{2}\,t_0-a\Big]^2=a^2. \tag{2}$$

Substitute $t_0$ from (1):

$$0.5\,t_0=\;0.5\left(\frac{\sqrt{3}}{2}\,a-0.5\,x_R\right)=\frac{\sqrt{3}}{4}\,a-0.25\,x_R,$$

and

$$\frac{\sqrt{3}}{2}\,t_0=\frac{\sqrt{3}}{2}\left(\frac{\sqrt{3}}{2}\,a-0.5\,x_R\right)=\frac{3}{4}\,a- \frac{\sqrt{3}}{4}\,x_R.$$

Thus, the coordinates of $Q$ become:

$$x_Q=x_R+ \frac{\sqrt{3}}{4}\,a-0.25\,x_R =0.75\,x_R+\frac{\sqrt{3}}{4}\,a,$$ $$y_Q=\frac{3}{4}\,a-\frac{\sqrt{3}}{4}\,x_R.$$

Plugging these in the sphere equation (2):

$$\left(0.75\,x_R+\frac{\sqrt{3}}{4}\,a\right)^2+\left[\left(\frac{3}{4}\,a-\frac{\sqrt{3}}{4}\,x_R\right)-a\right]^2=a^2.$$

Notice that

$$\left(\frac{3}{4}\,a-\frac{\sqrt{3}}{4}\,x_R-a\right)= -\frac{1}{4}\,a-\frac{\sqrt{3}}{4}\,x_R.$$

Expanding both squares (and using exact values $0.75=\frac{3}{4}$ and $\frac{\sqrt{3}}{4}$):

$$\left(\frac{3}{4}x_R+\frac{\sqrt{3}}{4}a\right)^2+ \left(\frac{1}{4}a+\frac{\sqrt{3}}{4}x_R\right)^2=a^2.$$

Expanding,

$$\frac{9}{16}x_R^2+\frac{2\cdot3\sqrt{3}}{16}a\,x_R+\frac{3}{16}a^2+\frac{1}{16}a^2+\frac{2\sqrt{3}}{16}a\,x_R+\frac{3}{16}x_R^2=a^2.$$

Combine like terms:

• Coefficient of $x_R^2$: $\frac{9}{16}+\frac{3}{16}=\frac{12}{16}=\frac{3}{4}$.
• Coefficient of $ax_R$: $\frac{6\sqrt{3}}{16}+\frac{2\sqrt{3}}{16}=\frac{8\sqrt{3}}{16}=\frac{\sqrt{3}}{2}$.
• Coefficient of $a^2$: $\frac{3}{16}+\frac{1}{16}=\frac{1}{4}$.

Thus the equation reduces to:

$$\frac{3}{4}x_R^2+\frac{\sqrt{3}}{2}a\,x_R+\frac{1}{4}a^2=a^2.$$

Multiply through by 4 to clear denominators:

$$3x_R^2+2\sqrt{3}\,a\,x_R+a^2=4a^2,$$

or

$$3x_R^2+2\sqrt{3}\,a\,x_R-3a^2=0.$$

Dividing by 3:

$$x_R^2+\frac{2\sqrt{3}}{3}\,a\,x_R-a^2=0. \tag{3}$$

Solve (3) using the quadratic formula:

$$x_R=\frac{-\frac{2\sqrt{3}}{3}a\pm\sqrt{\left(\frac{2\sqrt{3}}{3}a\right)^2+4a^2}}{2}.$$

Compute the discriminant:

$$\left(\frac{2\sqrt{3}}{3}a\right)^2+4a^2=\frac{4\cdot 3}{9}a^2+4a^2=\frac{4}{3}a^2+4a^2=\frac{16}{3}a^2.$$

Thus,

$$x_R=\frac{-\frac{2\sqrt{3}}{3}a\pm\frac{4}{\sqrt{3}}a}{2}.$$

Taking the positive root (since $x_R$ is taken as positive),

$$x_R=\frac{-\frac{2\sqrt{3}}{3}a+\frac{4a}{\sqrt{3}}}{2}=\frac{\left(\frac{-2\sqrt{3}+4\sqrt{3}}{3}\right)a}{2}=\frac{\frac{2\sqrt{3}}{3}a}{2}=\frac{a\sqrt{3}}{3}.$$

Now equate this result with our earlier relation (A):

$$\frac{a^2}{3}=\frac{n^2a^2}{4(4-n^2)}.$$

Cancel $a^2$ (with $a\neq0$) and cross–multiply:

$$4(4-n^2)=3n^2.$$

That is,

$$16-4n^2=3n^2\quad\Longrightarrow\quad16=7n^2,$$

so that,

$$n^2=\frac{16}{7}\quad\Longrightarrow\quad n=\frac{4}{\sqrt{7}}.$$

Thus, the refractive index of the liquid is $\displaystyle \frac{4}{\sqrt{7}}$ which is option D.

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Summary of the solution:

1. Coordinate Setup:

   • Interface at $y=0$, sphere with center $(0,a)$ and radius $a$; top touches $(0,0)$.
   • Point source taken at $S=(0,-a/2)$ (on the vertical through the sphere).

2. For the ray coming from $S$ and refracting at $R=(x_R,0)$:

   • By Snell’s law: $$\frac{|x_R|}{\sqrt{x_R^2+(a/2)^2}}=\frac{n}{2}.$$
   • This gives $x_R^2=\frac{n^2a^2}{4(4-n^2)}.$

3. Tangent Condition to the Sphere:

   • The ray (with direction $(0.5,\sqrt{3}/2)$) touches the sphere tangentially, producing a second equation which on solution gives $$x_R=\frac{a\sqrt{3}}{3}.$$

4. Combining the two results yields:

   $$n=\frac{4}{\sqrt{7}}.$$

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Final Answer: $\displaystyle \frac{4}{\sqrt{7}}$ (option D).

Subject: NCERT Physics – Optics
Chapter: Ray Optics
Topic: Refraction of Light