When you walk on a railway if we can see a straight rail road, we see the two rails meet together at the horizon. But yet we know that the two rails are parallel to each other. This is due to the perception of the eye. The gist of the post is not to analyze the above matter but a similar perception that is bit advanced. This came to my mind when I was travelling on the train. Most of you might have experienced the same phenomena. That is, when you look outside of a travelling train, we can see that the objects near to the train move faster, while the objects that is distant move slower. We know that the concept of movement of the 'objects outside' is explained using the relative motion, but why do they move at different speeds? Let's analyze this scenario in this post.
In this post I've assumed that the eye lobe is a perfect sphere. Lets consider a point that is moving parallel at a $l$ distant from the eye.
Here,
$d$ = diameter of the eye lobe
$x$ = distant of a moving point from the reference
$s$ = arc length between the image of the point and the reference
$$x = l\sin \alpha$$
When differentiated with respect to time,
$$\frac{dx}{dt} = l\cos\alpha \frac{d\alpha}{dt}\qquad (1)$$
But,
$$ s = \frac{d}{2} (2\alpha)=d\alpha$$
When differentiated with respect to time,
$$\begin{align}
\frac{ds}{dt} &= \frac{d(d\alpha)}{dt}\qquad \text{($d$ = constant)}\\
\frac{d\alpha}{dt} &= \frac 1d \frac{ds}{dt}\qquad (2)\\
\end{align}$$
From equation 2 and 1,
$$\frac{dx}{dt} = l\cos \alpha \frac 1d \frac{ds}{dt}$$
Since,
$$\begin{align}
\frac{dx}{dt} &= v\qquad \text{($v$ = velocity)}\\
v &= l\cos \alpha \frac 1d \frac{ds}{dt}\\
\frac{ds}{dt} &= \frac{vd}{l\cos\alpha} \qquad (3)
\end{align}$$
Although the actual velocity of the point is $v$, the velocity observed by the eye is denoted by the expression $\frac{ds}{dt}$ (lets call it 'visible velocity').
Consider two instances where both points move at same velocity,
Since it is mentioned that the both points move at the same velocity ($v$) and the diameter of the eye lobe doesn't change at two scenarios
If the respective visible velocities are, $\frac{ds_1}{dt}$ and $\frac{ds_2}{dt}$, From equation 3,
$$\frac{ds_1}{dt} = \frac{vd}{l_1 \cos \alpha_1}, \quad \frac{ds_2}{dt} = \frac{vd}{l_2\cos \alpha_2}$$
Since,
$$\alpha_1 >\alpha_2\implies \cos \alpha_1 < \cos \alpha_2$$
and
$$l_1< l_2$$
Therefore,
$$\begin{align}
l_1\cos\alpha_1 &<l_2\cos\alpha_2 \\
\frac{1}{l_1\cos\alpha_1} & > \frac{1}{l_2\cos\alpha_2}\\
\frac{vd}{l_1\cos\alpha_1} & > \frac{vd}{l_2\cos\alpha_2}\\
\frac{ds_1}{dt} &> \frac{ds_2}{dt}
\end{align}$$
So this means that the visible velocity of the closer object is larger than the visible velocity of the object which is far.
This proves that when the eye is moving parallel to two images, it recognizes that the image situated nearer to the eye moving faster than the image situated behind
