Ray Optics and Optical JEE 2022: Solved Advanced PYQ

Q: How many reflections occur before the ray returns through P?

For θ = 60° and 0 < l < L/2, the ray undergoes five reflections before returning through P. Its arrival at the hole is an emergence event, not a sixth reflection.

Q: Why is option D false in the JEE Advanced 2022 optics question?

Option D incorrectly counts emergence through P as a reflection from the base mirror. The ray has five mirror collisions and then exits through the hole, so the stated count of six reflections is wrong.

Q: Which formula should I use to trace reflections from inclined mirrors?

Use α′ = 2β − α, where α is the incoming direction angle and β is the mirror inclination from the positive x-axis. Interpret the result modulo 360°. Combine this rule with exact line intersections to find every collision point.

What is the Ray Optics and Optical JEE 2022 question?

The verified answer to the ray optics and optical JEE 2022 multiple-reflection question is A and B only. This hard, multi-correct question appeared in JEE Advanced 2022 Paper 1. Expect a solving time of about 240 seconds because checking all four options requires several exact intersections and a complete reflection count.

Three plane mirrors form an equilateral triangle ABC with side length denoted by capital L. A and B are the base corners, C is the top vertex, and a hole P lies on AB at distance lowercase l from B. A ray enters through P, moving up-left at angle theta with the base, and can emerge only through the same hole.

The four claims are:

A: For

θ = 30^{\circ}, 0 < l < L

the ray returns through P for every allowed position of the hole.

B: For $l = \frac{L}{2}$

there is an angle for which the ray returns through P after exactly two reflections.

C: For

θ = 60^{\circ}, l = \frac{L}{3}

the ray never returns through P.

D: For

θ = 60^{\circ}, 0 < l < \frac{L}{2}

the ray returns through P after six reflections.

Which coordinate and reflection framework should you use?

Use one fixed coordinate system and represent each ray by its direction angle from the positive x-axis. This method gives both the collision point and the outgoing direction after every reflection. It is more reliable than estimating successive paths from a sketch.

Take the vertices and the hole as

A = (0, 0), B = (L, 0), C = (\frac{L}{2}, \frac{\sqrt{3} L}{2})

$P = (L - l, 0)$ The three mirror equations are $A B : y = 0$ $A C : y = \sqrt{3} x$ $B C : y = \sqrt{3} (L - x)$

If a ray has direction angle alpha and strikes a mirror inclined at beta to the positive x-axis, its reflected direction is $α^{'} = 2 β - α$

Interpret the result modulo 360 degrees. Theta is measured with the base, but the entering ray moves up-left. Its direction angle is therefore $α = 180^{\circ} - θ$

Why are options A and B correct?

Option A is correct because the ray returns to P after one reflection for every permitted hole position. Option B is correct because the midpoint position allows a path with exactly two mirror collisions. The intersection coordinates prove both claims without relying on visual symmetry.

How does option A return for every allowed value of lowercase l?

The ray reflects once from AC and returns exactly to P. Set the entry angle to 30 degrees. The incoming direction is 150 degrees, and the ray through P has slope negative one by square root three. $θ = 30^{\circ}$

\tan 150^{\circ} = - \frac{1}{\sqrt{3}}

Its line through P is

y = - \frac{1}{\sqrt{3}} [x - (L - l)]

Intersect this ray with AC: $y = \sqrt{3} x$

The first reflection point is

Q = (\frac{L - l}{4}, \frac{\sqrt{3} (L - l)}{4})

The inclination of AC is 60 degrees. After reflection,

α^{'} = 2 (60^{\circ}) - 150^{\circ} = - 30^{\circ}

The reflected ray has slope $- \frac{1}{\sqrt{3}}$

Its equation through Q is

y - \frac{\sqrt{3} (L - l)}{4} = - \frac{1}{\sqrt{3}} (x - \frac{L - l}{4})

At the base, set the vertical coordinate equal to zero. The intersection is $x = L - l$

This is exactly the coordinate of P. Hence A is correct for every value in the stated range.

How does option B produce exactly two reflections?

Option B works by placing P at the midpoint and choosing an entry angle of 60 degrees. The ray reflects first from AC and then from BC before reaching P. There is no collision with the base mirror before emergence.

Set $l = \frac{L}{2}$

Therefore, $P = (\frac{L}{2}, 0)$

Choose $θ = 60^{\circ}$

The entering direction is 120 degrees. Its first intersection with AC is

(\frac{L}{4}, \frac{\sqrt{3} L}{4})

Reflection from the 60-degree mirror gives

α^{'} = 2 (60^{\circ}) - 120^{\circ} = 0^{\circ}

The ray travels horizontally and meets BC at

(\frac{3 L}{4}, \frac{\sqrt{3} L}{4})

The right mirror has inclination 120 degrees. The reflected direction is

α^{'} = 2 (120^{\circ}) - 0^{\circ} = 240^{\circ}

This ray reaches the base at $(\frac{L}{2}, 0) = P$

The required angle exists, and the ray returns after exactly two reflections. B is correct.

Why are options C and D false?

Both claims fail because the 60-degree ray returns to P after five reflections, not zero or six. The first return to the base is not an emergence unless it occurs at P. Following every collision coordinate settles both the return point and the reflection count.

Start with

P = (L - l, 0), 0 < l < \frac{L}{2}

The initial direction is $α = 120^{\circ}$

The ray first meets AC at

(\frac{L - l}{2}, \frac{\sqrt{3} (L - l)}{2})

Reflection from AC changes its direction to

α^{'} = 2 (60^{\circ}) - 120^{\circ} = 0^{\circ}

The horizontal ray meets BC at

(\frac{L + l}{2}, \frac{\sqrt{3} (L - l)}{2})

Reflection from BC changes its direction to

α^{'} = 2 (120^{\circ}) - 0^{\circ} = 240^{\circ}

The ray then meets AB at $(l, 0)$

Reflection from the base changes its direction to 120 degrees. It next meets AC at

(\frac{l}{2}, \frac{\sqrt{3} l}{2})

Reflection from AC changes the direction to zero degrees. The horizontal ray meets BC at

(L - \frac{l}{2}, \frac{\sqrt{3} l}{2})

Reflection from BC changes the direction to 240 degrees. The final ray travels down-left and reaches the base at $x = L - l$

This is the original hole P.

For option C, substitute $l = \frac{L}{3}$

The ray first reaches the base at $x = \frac{L}{3}$

The hole is at

x = L - l = \frac{2 L}{3}

The first base point is not the hole. The ray reflects there, continues through two more mirror collisions, and later emerges through P. Therefore C is false.

For option D, the events occur in this order:

Reflection from the left mirror
Reflection from the right mirror
Reflection from the base mirror
Reflection from the left mirror
Reflection from the right mirror
Emergence through P

Only the first five events are reflections. Therefore D is false.

Verified final answer: A and B only.

What method error produces the six-reflection answer?

The six-reflection answer comes from counting arrival at P as another collision with the base mirror. A reflection occurs only when the ray strikes a reflecting surface and remains inside the enclosure. At P, the ray passes through the opening and emerges.

The correct event count is:

Five mirror collisions
One emergence event
Five reflections, not six reflections

Stopping the trace at the first return to the base is a different error. That collision occurs at $x = l$

The hole is at $x = L - l$

These points are equal only when $l = \frac{L}{2}$

For other values, the ray reflects from the first base point and continues inside the enclosure.

Which related ray optics questions should you practise?

Practise the direction-angle rule, image counting and reflection counting. These three questions test the same operations used in the ray optics and optical JEE 2022 problem: transforming a direction, applying the plane-mirror image formula and separating a reflection from emergence.

What is the final direction after reflections from 60-degree and 120-degree mirrors?

The final direction is the initial direction plus 120 degrees, interpreted modulo 360 degrees. Apply the reflection rule once at each mirror rather than trying to estimate the final ray from a sketch.

A ray with direction alpha reflects first from a mirror inclined at 60 degrees and then from one inclined at 120 degrees.

After the first reflection, $α_{1} = 120^{\circ} - α$

After the second, $α_{2} = 240^{\circ} - α_{1}$

Therefore,

α_{2} = α + 120^{\circ} (\mod 360^{\circ})

How many images do two mirrors inclined at 60 degrees form?

The two mirrors form five images when the object lies on their angle bisector. Divide 360 degrees by the angle between the mirrors, then subtract one.

\frac{360^{\circ}}{60^{\circ}} - 1 = 5

Hence the answer is five images.

Where does the ray first hit the base, and how many reflections occur?

The other base point has horizontal coordinate lowercase l, and the ray undergoes five reflections before returning through P. The collision at the other base point is the third reflection, not an emergence event.

For the same equilateral enclosure, take a 60-degree ray starting from

P = (L - l, 0), 0 < l < \frac{L}{2}

The other base collision occurs at

x = l

The ray then continues and returns through P after

5 reflections

For any closed mirror path, list the collision coordinates in order and classify the final event as either reflection or emergence.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Frequently asked questions

What is the answer to the Ray Optics JEE Advanced 2022 question?

The verified answer is A and B only. Option A returns the ray through P after one reflection for every allowed value of l. Option B gives exactly two reflections when l = L/2 and θ = 60°.

How many reflections occur before the ray returns through P?