A ray of light is projected from the origin at angle of `60^@` with the positive direction of x-axis towards the line, y = 2, which gets reflected from the point (`alpha`,2). Then the distance of the reflected ray of light from the point (2, 2) is:

Options

A. `3-sqrt3`
B. `sqrt3-1`
C. `1-1//sqrt3`
D. `2(1-1//sqrt3)`

Answer

To solve the problem step by step, we will follow the process outlined in the video transcript. Here is the detailed solution:

### Step 1: Understand the problem
We have a ray of light projected from the origin (0, 0) at an angle of \(60^\circ\) with the positive x-axis towards the line \(y = 2\). The ray reflects off the line at a point \((\alpha, 2)\). We need to find the distance of the reflected ray from the point (2, 2).

### Step 2: Determine the equation of the incident ray
The slope of the ray making an angle of \(60^\circ\) with the x-axis is given by:
\[ m = \tan(60^\circ) = \sqrt{3} \]
The equation of the line (ray of light) passing through the origin is:
\[ y = mx \implies y = \sqrt{3}x \]

### Step 3: Find the intersection point with the line \(y = 2\)
To find the intersection point of the ray with the line \(y = 2\), we set:
\[ \sqrt{3}x = 2 \implies x = \frac{2}{\sqrt{3}} \implies \alpha = \frac{2}{\sqrt{3}} \]
Thus, the intersection point is:
\[ \left(\frac{2}{\sqrt{3}}, 2\right) \]

### Step 4: Find the slope of the reflected ray
The angle of incidence is equal to the angle of reflection. The angle of the incident ray with the positive x-axis is \(60^\circ\). Therefore, the angle of the reflected ray with the positive x-axis is:
\[ 180^\circ - 60^\circ = 120^\circ \]
The slope of the reflected ray is:
\[ m' = \tan(120^\circ) = -\sqrt{3} \]

### Step 5: Write the equation of the reflected ray
Using the point-slope form of the line, the equation of the reflected ray passing through the point \(\left(\frac{2}{\sqrt{3}}, 2\right)\) is:
\[ y - 2 = -\sqrt{3}\left(x - \frac{2}{\sqrt{3}}\right) \]
Expanding this gives:
\[ y - 2 = -\sqrt{3}x + 2 \implies y = -\sqrt{3}x + 4 \]

### Step 6: Find the distance from the point (2, 2) to the reflected ray
The distance \(D\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is given by:
\[ D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
First, we rewrite the equation of the reflected ray in the standard form:
\[ \sqrt{3}x + y - 4 = 0 \]
Here, \(A = \sqrt{3}\), \(B = 1\), and \(C = -4\). Now, substituting the point (2, 2):
\[ D = \frac{|\sqrt{3}(2) + 1(2) - 4|}{\sqrt{(\sqrt{3})^2 + 1^2}} = \frac{|2\sqrt{3} + 2 - 4|}{\sqrt{3 + 1}} = \frac{|2\sqrt{3} - 2|}{2} \]
This simplifies to:
\[ D = |\sqrt{3} - 1| \]

### Final Answer
Thus, the distance of the reflected ray from the point (2, 2) is:
\[ \sqrt{3} - 1 \]