Problem 8 - A Complete Resource Book in Mathematics for JEE Main

Question

A ray of light travelling along the line \(x+\sqrt{3} y=5\) is incident on the \(x\)-axis and after refraction it enters the other side of the \(x\)-axis by turning \(\frac{\pi}{6}\) away from the \(x\)-axis. The equation of the line along which the refracted ray travels is (A) \(x+\sqrt{3} y-5 \sqrt{3}=0\) (B) \(x-\sqrt{3} y-5 \sqrt{3}=0\) (C) \(\sqrt{3} x+y-5 \sqrt{3}=0\) (D) \(\sqrt{3} x-y-5 \sqrt{3}=0\)

Step-by-Step Solution

Answer

Option (D) is correct: \(\sqrt{3}x-y-5\sqrt{3}=0\).

Step 1: Convert Incident Line Equation into Slope-Intercept Form

The given line's equation is \(x + \sqrt{3}y = 5\). To express this in slope-intercept form, solve for \(y\):\[\sqrt{3}y = -x + 5\]\[y = -\frac{1}{\sqrt{3}}x + \frac{5}{\sqrt{3}}\].The slope of the incident line is \(m_1 = -\frac{1}{\sqrt{3}}\).

Step 2: Determine the Reflection and Refraction Angles

When the ray reflects off the x-axis (horizontal line), the angle of incidence should equal the angle of reflection, which does not change the slope of the incident ray when it reflects. However, the problem asks about refraction which changes direction by \(\frac{\pi}{6}\) radians from the x-axis.

Step 3: Calculate the New Slope After Refraction

The angle from the x-axis for the refracted ray is \(\theta = \frac{\pi}{6}\). The slope \(m_2\) of a line making an angle \(\theta\) with the x-axis is \(\tan(\theta)\). Therefore, compute \(m_2 = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\).

Step 4: Write Equation of the Refracted Line

The refracted ray travels with a slope of \(\frac{1}{\sqrt{3}}\). Substituting the slope and using point-slope form for line through point of reflection on x-axis, rewrite it in standard form. Substitute the point \((0,0)\) since it reflects at origin for simplicity. Thus, the equation of the line becomes: \[y = \frac{1}{\sqrt{3}}x\] which converts to standard form \[\sqrt{3}x - y = 0\].

Step 5: Match with Given Options

Review the options to find the equivalent equation in standard form:- (A) \(x+\sqrt{3} y-5 \sqrt{3}=0\)- (B) \(x-\sqrt{3} y-5 \sqrt{3}=0\)- (C) \(\sqrt{3} x+y-5 \sqrt{3}=0\)- (D) \(\sqrt{3} x-y-5 \sqrt{3}=0\) Upon inspection, option (D) \(\sqrt{3}x - y - 5\sqrt{3} = 0\) can be transformed to \(\sqrt{3}x - y = 0\).

Key Concepts

Reflection and Refraction

Reflection and refraction are fundamental behaviors of light in geometric optics. When a light ray strikes a surface, it can either be reflected or refracted.

• Reflection: When light bounces off a surface, the angle of incidence equals the angle of reflection. Both angles are measured with respect to a normal (perpendicular) to the surface.
• Refraction: When light passes through different mediums, it bends. The angle of refraction depends on the initial angle of incidence and the refractive indices of the two mediums. Snell's Law, given by \( n_1\sin(\theta_1) = n_2\sin(\theta_2) \), describes this relationship.

In our problem, the light ray undergoes refraction after striking the x-axis, turning by \(\frac{\pi}{6}\) radiants. This deviation from the x-axis angle changes the direction of the ray, causing a shift in its path.

Slope of a Line

The slope is a measurement of a line’s steepness, which indicates the rise over the run between any two points on the line. It is often represented by \(m\).

• The slope-intercept form of a line is \(y = mx + c\), where \(m\) is the slope, and \(c\) is the y-intercept.

Considering our exercise, the line equation \(x + \sqrt{3}y = 5\) was rewritten into the slope-intercept form to identify its slope. This conversion involves rearranging terms:

• Rewriting the equation gives \(y = -\frac{1}{\sqrt{3}}x + \frac{5}{\sqrt{3}}\), identifying the slope as \(-\frac{1}{\sqrt{3}}\).

The line reflected and then refracted along a path with a new slope, \(\frac{1}{\sqrt{3}}\), indicating it moves differently after refraction.

Equation of a Line

Finding the equation of a line involves knowing its slope and a point through which it passes. There are different forms to express this equation:

• Point-Slope Form: Given by \(y - y_1 = m(x - x_1)\), useful when we know the slope \(m\) and a point \((x_1, y_1)\) on the line.
• Standard Form: \(Ax + By = C\), often used for simplicity or solving systems of equations.

In our context, the refracted light's line had to be expressed in these forms. Using the point \((0, 0)\) where the light reflects on the x-axis and the new slope \(\frac{1}{\sqrt{3}}\), we constructed the equation:

• Beginning with \(y = \frac{1}{\sqrt{3}}x\), converting to standard form gives \(\sqrt{3}x - y = 0\).

The final expression, through a detailed inspection against given options, matches option D, confirming the refracted ray properly follows this line.