A beam of light is sent along the line `x-y=1` , which after refracting from the x-axis enters the opposite side by turning through `30^0` towards the normal at the point of incidence on the x-axis. Then the equation of the refracted ray is `(2-sqrt(3))x-y=2+sqrt(3)` `(2+sqrt(3))x-y=2+sqrt(3)` `(2-sqrt(3))x+y=(2+sqrt(3))` `y=(2-sqrt(3))(x-1)`

Options

A. `(2-sqrt(3))x-y=2+sqrt(3)`
B. `(2+sqrt(3))x-y=2+sqrt(3)`
C. `(2-sqrt(3))x+y=(2+sqrt(3))`
D. `y=(2+sqrt(3))(x-1)`

Explanation

To solve the problem step by step, we need to find the equation of the refracted ray after the beam of light refracts off the x-axis. Here’s how we can do it:

### Step 1: Understand the given line
The line given is \( x - y = 1 \). We can rewrite this in slope-intercept form:
\[ y = x - 1 \]
This line has a slope of 1, which means it makes an angle of \( 45^\circ \) with the positive x-axis (since \( \tan(45^\circ) = 1 \)).

**Hint:** Convert the line equation to slope-intercept form to easily identify the slope.

### Step 2: Determine the angle of incidence
Since the line makes an angle of \( 45^\circ \) with the positive x-axis, the angle of incidence when the light hits the x-axis is also \( 45^\circ \).

**Hint:** Remember that the angle of incidence is the angle between the incoming ray and the normal to the surface at the point of incidence.

### Step 3: Calculate the angle of refraction
The light refracts at the x-axis and turns through \( 30^\circ \) towards the normal. Therefore, the angle of refraction will be:
\[ \text{Angle of refraction} = 45^\circ - 30^\circ = 15^\circ \]

**Hint:** Subtract the angle turned from the angle of incidence to find the angle of refraction.

### Step 4: Find the slope of the refracted ray
The slope of the refracted ray can be found using the tangent of the angle of refraction:
\[ \text{slope} = \tan(15^\circ) \]
Using the tangent subtraction formula:
\[ \tan(15^\circ) = \frac{\tan(45^\circ) - \tan(30^\circ)}{1 + \tan(45^\circ) \tan(30^\circ)} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]
This simplifies to:
\[ \tan(15^\circ) = 2 - \sqrt{3} \]

**Hint:** Use the tangent subtraction formula to calculate the tangent of the angle.

### Step 5: Write the equation of the refracted ray
The refracted ray passes through the point of incidence on the x-axis, which is \( (1, 0) \). We can use the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting \( y_1 = 0 \), \( m = 2 - \sqrt{3} \), and \( x_1 = 1 \):
\[ y - 0 = (2 - \sqrt{3})(x - 1) \]
This simplifies to:
\[ y = (2 - \sqrt{3})(x - 1) \]

**Hint:** Use the point-slope form of a line to find the equation when you know a point and the slope.

### Final Answer
The equation of the refracted ray is:
\[ y = (2 - \sqrt{3})(x - 1) \]