Question

A ray parallel to diameter $AB$ of spherical drop passes through point 'A' after refracting from the drop, then refractive index of liquid of drop is___

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. 2
• D. 1

Answer 1

Answer: (D)

1

Answer 2

Let the radius of the spherical drop be $R$. Let the center of the sphere be $O$. Let the diameter be $AB$. The incident ray is parallel to $AB$. Let this ray be at a height $h$ from $AB$. The ray enters the drop at point $P$. The normal to the surface at $P$ is the radius $OP$. Let $\phi$ be the angle between $OP$ and the diameter $AB$. Then $h = R \sin \phi$. The angle of incidence $i$ is the angle between the incident ray and the normal $OP$. Since the incident ray is parallel to $AB$, the angle between the incident ray and $OP$ is $i = 90^\circ - \phi$. The problem states that this angle is $\theta$. Thus, $i = \theta$, which implies $\theta = 90^\circ - \phi$, or $\phi = 90^\circ - \theta$.

According to Snell's law at point $P$, $1 \cdot \sin i = \mu \sin r$, where $\mu$ is the refractive index of the liquid and $r$ is the angle of refraction. So, $\sin \theta = \mu \sin r$.

The ray refracts and travels inside the sphere. The problem states that the ray passes through point $A$ after refracting from the drop. The figure shows that the ray exits the drop at point $A$. So, the point of exit $Q$ is $A$. Point $A$ lies on the diameter $AB$. The normal at $A$ is the radius $OA$. The ray inside the sphere is the line segment $PA$. The angle of incidence at $A$ is $i'$, which is the angle between the ray $PA$ and the normal $OA$. The angle of refraction at $A$ is $r'$, which is the angle between the emergent ray and the normal $OA$.

In triangle $OPA$, $OP = OA = R$, so it is an isosceles triangle. The angle $\angle POA$ is the angle between the radii $OP$ and $OA$. Since $OA$ lies along the diameter $AB$, and $\phi$ is the angle between $OP$ and $AB$, we have $\angle POA = \phi = 90^\circ - \theta$. In isosceles triangle $OPA$, $\angle OAP = \angle OPA = \frac{180^\circ - \angle POA}{2} = \frac{180^\circ - (90^\circ - \theta)}{2} = \frac{90^\circ + \theta}{2} = 45^\circ + \frac{\theta}{2}$. The angle of incidence at $A$ is $i' = \angle OAP = 45^\circ + \frac{\theta}{2}$. The angle of refraction at $P$ is $r$, which is the angle between $OP$ and $PA$. So, $r = \angle OPA = 45^\circ + \frac{\theta}{2}$. Thus, $r = i'$.

Now, applying Snell's law at $A$: $\mu \sin i' = 1 \cdot \sin r'$. Since the ray exits at $A$, the emergent ray is the extension of the ray $PA$. Therefore, the angle of refraction at $A$, $r'$, is the angle between the emergent ray (line $PA$) and the normal $OA$. This is precisely the angle $i'$. So, $r' = i'$. Substituting this into Snell's law at $A$: $\mu \sin i' = \sin i'$. Since $i' = 45^\circ + \theta/2$, and $\theta$ is typically a small angle for such phenomena, $\sin i' \neq 0$. Thus, this equation implies $\mu = 1$.