JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Finding the $x$ -intercept of a reflected light ray

Q: Why is the reflected inclination θ_r = 2θ_m - θ_i?

The mirror makes angle θ_m with the x -axis. The incident angle to the mirror is θ_m - θ_i. By symmetry, the reflected line makes the same angle on the other side, so its angle with the positive x -axis is θ_m + (θ_m - θ_i) = 2θ_m - θ_i.

A light ray emits from the origin making an angle 30° with the positive x -axis. After getting reflected by the line x + y = 1, if this ray intersects x -axis at Q, then the abscissa of Q is

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A(√(3))/(2(√(3)+1))B2/(3+√(3))C2/((√(3)-1))D2/(3-√(3))

Step-by-step solutionView

Correct answer

The abscissa of the point where the reflected ray meets the $x$ -axis is $\frac{2}{3 + \sqrt{3}}$ .

Option analysis

Why each option works or fails

A · (√(3))/(2(√(3)+1))

Confusing the coordinates of the point of incidence on $x + y = 1$ with the abscissa of the intercept $Q$ on the $x$ -axis. After finding the point of incidence $P$ , write the equation of the reflected ray and solve for its $x$ -intercept by setting $y = 0$ .

B · 2/(3+√(3))

This is the correct option. The ray strikes $x + y = 1$ at $(\frac{\sqrt{3}}{\sqrt{3} + 1}, \frac{1}{\sqrt{3} + 1})$ and reflects at an angle of inclination of $240^{\circ}$ (or slope $\sqrt{3}$ ), yielding an $x$ -intercept of $\frac{2}{3 + \sqrt{3}}$ .

C · 2/((√(3)-1))

Sign error when calculating the mirror's reflection angle or during the intersection equation with the line $x + y = 1$ . Ensure the incident ray $y = \frac{1}{\sqrt{3}} x$ satisfies $x + y = 1$ yielding $x = \frac{\sqrt{3}}{\sqrt{3} + 1}$ , not $\frac{\sqrt{3}}{\sqrt{3} - 1}$ .

D · 2/(3-√(3))

Using the wrong sign for the slope of the reflected ray, taking $m_{ref} = - \sqrt{3}$ instead of $+ \sqrt{3}$ . The inclination of the mirror is $135^{\circ}$ . The reflected inclination is $2 (135^{\circ}) - 30^{\circ} = 240^{\circ}$ , giving slope $\tan (240^{\circ}) = \sqrt{3}$ .

Reviewed route

Solution

StepWorking

01Given

Incident ray leaves the origin $O (0, 0)$ with inclination $30^{\circ}$ , reflecting off the mirror line $x + y = 1$ , and its reflected ray intersects the $x$ -axis at $Q (x_{Q}, 0)$ .

02Goal

Find the abscissa $x_{Q}$ of the point $Q$ .

03Approach

The line containing the reflected ray passes through the point of incidence $P$ on $x + y = 1$ and the image of any point on the incident ray across the mirror line. Taking the image of $O (0, 0)$ gives an easy second point $M (1, 1)$ on the reflected ray line. Finding the intersection $P$ of the incident ray $y = x / \sqrt{3}$ with $x + y = 1$ determines the reflected ray line, from which setting $y = 0$ gives $x_{Q}$ .

04Execute

Find the image of $O (0, 0)$ in $x + y - 1 = 0$ : $\frac{x - 0}{1} = \frac{y - 0}{1} = - 2 \frac{0 + 0 - 1}{1^{2} + 1^{2}} = 1 ⟹ M (1, 1)$ Now find intersection $P$ of $y = \frac{x}{\sqrt{3}}$ and $x + y = 1$ : $x + \frac{x}{\sqrt{3}} = 1 ⟹ x \frac{\sqrt{3} + 1}{\sqrt{3}} = 1 ⟹ x = \frac{\sqrt{3}}{\sqrt{3} + 1} = \frac{3 - \sqrt{3}}{2}$ $y = 1 - x = \frac{\sqrt{3} - 1}{2}$ Slope of the reflected ray line $P M$ is: $m = \frac{1 - \frac{\sqrt{3} - 1}{2}}{1 - \frac{3 - \sqrt{3}}{2}} = \frac{3 - \sqrt{3}}{\sqrt{3} - 1} = \sqrt{3}$ Equation of the reflected line through $M (1, 1)$ : $y - 1 = \sqrt{3} (x - 1) ⟹ y = \sqrt{3} x + (1 - \sqrt{3})$ Setting $y = 0$ to find the $x$ -intercept $Q (x_{Q}, 0)$ : $0 = \sqrt{3} x_{Q} + 1 - \sqrt{3} ⟹ x_{Q} = \frac{\sqrt{3} - 1}{\sqrt{3}} = 1 - \frac{1}{\sqrt{3}}$ Rationalizing in the form matching the options: $x_{Q} = \frac{(\sqrt{3} - 1) (\sqrt{3} + 1)}{\sqrt{3} (\sqrt{3} + 1)} = \frac{3 - 1}{3 + \sqrt{3}} = \frac{2}{3 + \sqrt{3}}$

✓Verify

Check angle of reflection directly: mirror line inclination is $135^{\circ}$ . Incident ray inclination is $30^{\circ}$ . Angle of incidence with mirror = $135^{\circ} - 30^{\circ} = 105^{\circ}$ (or $75^{\circ}$ acute). Reflected ray angle of inclination $θ_{r}$ : since reflection across line with inclination $135^{\circ}$ maps angle $θ$ to $2 (135^{\circ}) - θ = 270^{\circ} - 30^{\circ} = 240^{\circ} \equiv 60^{\circ}$ . Slope of reflected ray is $\tan 60^{\circ} = \sqrt{3}$ . This matches our calculated slope $m = \sqrt{3}$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why does the image of the origin lie on the line containing the reflected ray?

By the law of reflection, the reflected ray appears to emanate from the virtual image of the light source. Hence, the line of the reflected ray passes through the mirror image of any point on the incident ray.

Why is the reflected inclination

θ_{r} = 2 θ_{m} - θ_{i}

The mirror makes angle $θ_{m}$ with the $x$ -axis. The incident angle to the mirror is $θ_{m} - θ_{i}$ . By symmetry, the reflected line makes the same angle on the other side, so its angle with the positive $x$ -axis is $θ_{m} + (θ_{m} - θ_{i}) = 2 θ_{m} - θ_{i}$ .

Answer

The abscissa of the point where the reflected ray meets the $x$ -axis is $\frac{2}{3 + \sqrt{3}}$ .

Why each option works or fails

A: (√(3))/(2(√(3)+1)) - Confusing the coordinates of the point of incidence on $x + y = 1$ with the abscissa of the intercept $Q$ on the $x$ -axis. After finding the point of incidence $P$ , write the equation of the reflected ray and solve for its $x$ -intercept by setting $y = 0$ .
B · correct: 2/(3+√(3)) - This is the correct option. The ray strikes $x + y = 1$ at $(\frac{\sqrt{3}}{\sqrt{3} + 1}, \frac{1}{\sqrt{3} + 1})$ and reflects at an angle of inclination of $240^{\circ}$ (or slope $\sqrt{3}$ ), yielding an $x$ -intercept of $\frac{2}{3 + \sqrt{3}}$ .
C: 2/((√(3)-1)) - Sign error when calculating the mirror's reflection angle or during the intersection equation with the line $x + y = 1$ . Ensure the incident ray $y = \frac{1}{\sqrt{3}} x$ satisfies $x + y = 1$ yielding $x = \frac{\sqrt{3}}{\sqrt{3} + 1}$ , not $\frac{\sqrt{3}}{\sqrt{3} - 1}$ .
D: 2/(3-√(3)) - Using the wrong sign for the slope of the reflected ray, taking $m_{ref} = - \sqrt{3}$ instead of $+ \sqrt{3}$ . The inclination of the mirror is $135^{\circ}$ . The reflected inclination is $2 (135^{\circ}) - 30^{\circ} = 240^{\circ}$ , giving slope $\tan (240^{\circ}) = \sqrt{3}$ .

Step-by-step solution

Given: Incident ray leaves the origin $O (0, 0)$ with inclination $30^{\circ}$ , reflecting off the mirror line $x + y = 1$ , and its reflected ray intersects the $x$ -axis at $Q (x_{Q}, 0)$ .
Goal: Find the abscissa $x_{Q}$ of the point $Q$ .
Approach: The line containing the reflected ray passes through the point of incidence $P$ on $x + y = 1$ and the image of any point on the incident ray across the mirror line. Taking the image of $O (0, 0)$ gives an easy second point $M (1, 1)$ on the reflected ray line. Finding the intersection $P$ of the incident ray $y = x / \sqrt{3}$ with $x + y = 1$ determines the reflected ray line, from which setting $y = 0$ gives $x_{Q}$ .
Execute: Find the image of $O (0, 0)$ in $x + y - 1 = 0$ : $\frac{x - 0}{1} = \frac{y - 0}{1} = - 2 \frac{0 + 0 - 1}{1^{2} + 1^{2}} = 1 ⟹ M (1, 1)$ Now find intersection $P$ of $y = \frac{x}{\sqrt{3}}$ and $x + y = 1$ : $x + \frac{x}{\sqrt{3}} = 1 ⟹ x \frac{\sqrt{3} + 1}{\sqrt{3}} = 1 ⟹ x = \frac{\sqrt{3}}{\sqrt{3} + 1} = \frac{3 - \sqrt{3}}{2}$ $y = 1 - x = \frac{\sqrt{3} - 1}{2}$ Slope of the reflected ray line $P M$ is: $m = \frac{1 - \frac{\sqrt{3} - 1}{2}}{1 - \frac{3 - \sqrt{3}}{2}} = \frac{3 - \sqrt{3}}{\sqrt{3} - 1} = \sqrt{3}$ Equation of the reflected line through $M (1, 1)$ : $y - 1 = \sqrt{3} (x - 1) ⟹ y = \sqrt{3} x + (1 - \sqrt{3})$ Setting $y = 0$ to find the $x$ -intercept $Q (x_{Q}, 0)$ : $0 = \sqrt{3} x_{Q} + 1 - \sqrt{3} ⟹ x_{Q} = \frac{\sqrt{3} - 1}{\sqrt{3}} = 1 - \frac{1}{\sqrt{3}}$ Rationalizing in the form matching the options: $x_{Q} = \frac{(\sqrt{3} - 1) (\sqrt{3} + 1)}{\sqrt{3} (\sqrt{3} + 1)} = \frac{3 - 1}{3 + \sqrt{3}} = \frac{2}{3 + \sqrt{3}}$
Verify: Check angle of reflection directly: mirror line inclination is $135^{\circ}$ . Incident ray inclination is $30^{\circ}$ . Angle of incidence with mirror = $135^{\circ} - 30^{\circ} = 105^{\circ}$ (or $75^{\circ}$ acute). Reflected ray angle of inclination $θ_{r}$ : since reflection across line with inclination $135^{\circ}$ maps angle $θ$ to $2 (135^{\circ}) - θ = 270^{\circ} - 30^{\circ} = 240^{\circ} \equiv 60^{\circ}$ . Slope of reflected ray is $\tan 60^{\circ} = \sqrt{3}$ . This matches our calculated slope $m = \sqrt{3}$ .

Shortcut:

When to use it: Fastest method in JEE: use inclination angles directly to get the slope of the reflected ray, bypass finding the image point.

Given: Incident ray has inclination $θ_{i} = 30^{\circ}$ , mirror line $x + y = 1$ has inclination $θ_{m} = 135^{\circ}$ .

Goal: Find the reflected ray inclination directly, then find the intersection with the $x$ -axis.

Approach: Under reflection in a line of inclination $θ_{m}$ , the inclination of the reflected line is $θ_{r} = 2 θ_{m} - θ_{i} = 2 (135^{\circ}) - 30^{\circ} = 240^{\circ}$ , which gives slope $m_{r} = \tan 240^{\circ} = \sqrt{3}$ .

Execute: The point of incidence $P$ satisfies $x + \frac{x}{\sqrt{3}} = 1 ⟹ x_{P} = \frac{\sqrt{3}}{\sqrt{3} + 1}$ , $y_{P} = \frac{1}{\sqrt{3} + 1}$ . The reflected ray equation is $y - y_{P} = \sqrt{3} (x - x_{P})$ . Setting $y = 0$ : $- y_{P} = \sqrt{3} (x_{Q} - x_{P}) ⟹ x_{Q} = x_{P} - \frac{y_{P}}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3} + 1} - \frac{1}{\sqrt{3} (\sqrt{3} + 1)} = \frac{3 - 1}{\sqrt{3} (\sqrt{3} + 1)} = \frac{2}{3 + \sqrt{3}}$

Verify: Evaluate numerically: $2 / (3 + 1.732) = 2 / 4.732 \approx 0.423$ . From $x_{Q} = 1 - 1 / \sqrt{3} = 1 - 0.577 = 0.423$ . Matches precisely.

Finding the $x$ -intercept of a reflected light ray

What feels right?

The abscissa of the point where the reflected ray meets the $x$ -axis is $\frac{2}{3 + \sqrt{3}}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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