- #1
Efeguleroglu
- 24
- 2
(Black one is the object and grey is its image.)
We know from Snell's Law:
$$
n_1\sin\alpha=n_2\sin\theta
$$
And I have been said that:
$$
a=b\ (1)\\\ and\\\ \frac{h}{h'}=\frac{n_2}{n_1} \ (2)
$$
Let's begin.
$$
\frac{n_1}{n_2}=\frac{\sin\alpha}{\sin\theta}=\frac{\sin(\frac{\pi}{2}-\alpha)}{\cos(\frac{\pi}{2}-\theta)}=\frac{b}{\sqrt{b^2+h'^2}} \frac{\sqrt{a^2+h^2}}{a}
$$
Since$$a=b$$
$$\frac{n_2}{n_1}=\frac{\sqrt{a^2+h^2}}{\sqrt{a^2+h'^2}}$$
And also $$\frac{n_2}{n_1}=\frac{h}{h'}$$
Thus $$\frac{h}{h'}=\frac{\sqrt{a^2+h^2}}{\sqrt{a^2+h'^2}}$$
But it's true only if $$h=h'$$
But that means $$n_1=n_2$$
And we're not observing this.
So which one is wrong? [(1) or (2)] (maybe not wrong but assumption)