Question

A light ray parallel to the $x-$axis strikes the outer reflecting surface of a sphere at a point $(2,2,0)$. Its centre is at the point $(0,0,–1)$. The unit vector along the direction of the reflected ray is $x\hat{i}+y\hat{j}+z\hat{k}$. Find the value of $\frac{yz}{x^2}$.

Answer 1

Direction of incident ray, $\hat{e}=-\hat{i}$

Direction of normal ray, $\hat{n}=\frac{2\hat{i}+2\hat{j}+\hat{k}}{3}$

So, direction of reflected ray,

$\hat{r}=\hat{e}-2(\hat{e}.\hat{n})\hat{n}$

$\Rightarrow \hat{r}=(-\hat{i})-2\left[\left(\frac{-2}{3}\right)\frac{2\hat{i}+2\hat{j}+\hat{k}}{3}\right]$

$\Rightarrow \hat{r}=(-\hat{i})+\frac{4}{9}(2\hat{i}+2\hat{j}+\hat{k})$

$\Rightarrow \hat{r}=\frac{-\hat{i}+8\hat{j}+4\hat{k}}{9}$

According to the problem, the unit vector along the direction of reflected ray is $x\hat{i}+y\hat{j}+z\hat{k}$.

Now, on comparisons of two equations, we get the values of
$x,y$ and $z$ as $-1/9,8/9$ and $4/9$.

So, $\frac{yz}{x^2}=\frac{8/9\times 4/9}{(-1/9{)}^2}=32$