JEE MainPhysics

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+4 marks−1 if incorrectNumericalPrevious-year question

Refraction of Light and Shadow Formation at an Air-Water Interface

A pole is vertically submerged in swimming pool, such that it gives a length of shadow 2.15m within water when sunlight is incident at an angle of 30° with the surface of water. If swimming pool is filled to a height of 1.5m, then the height of the pole above the water surface in centimetres is ______. (n_()w = 4/3)

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Step-by-step solutionView

Correct answer

The height of the pole above the water surface is 50 cm.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Refractive index of water $n_{w} = \frac{4}{3}$ , angle of incident sunlight with water surface $θ = 30^{\circ}$ (so angle of incidence $i = 90^{\circ} - 30^{\circ} = 60^{\circ}$ ), depth of pool $d = 1.5 m$ , total length of shadow at the bottom $L = 2.15 m$ .

02Find

Height of the pole above the water surface $x$ in centimetres.

03Visualise

The top of the pole casts a ray entering water at a horizontal distance $x \tan i = x \tan 60^{\circ} = x \sqrt{3}$ from the pole's vertical line. Inside the water, the ray refracts at angle $r$ to the normal and traverses a depth of $1.5 m$ , adding a horizontal shift of $1.5 \tan r$ . The total shadow length on the bottom is $L = x \sqrt{3} + 1.5 \tan r = 2.15 m$ .

04Strategise

Use Snell's law $1 \cdot \sin 60^{\circ} = \frac{4}{3} \sin r$ to find $\sin r$ and $\tan r$ , then solve the linear geometric equation for $x$ .

05Execute

From Snell's law: $\sin r = \frac{3 \sqrt{3}}{8} \approx 0.6495$ . Then $\tan r = \frac{\sin r}{\sqrt{1 - \sin^{2} r}} = \frac{3 \sqrt{3}}{\sqrt{64 - 27}} = \frac{3 \sqrt{3}}{\sqrt{37}} \approx \frac{5.196}{6.0827} \approx 0.8542$ . Now solve for $x$ : $x \sqrt{3} + 1.5 (0.8542) = 2.15 ⟹ x \sqrt{3} = 2.15 - 1.2813 = 0.8687 ⟹ x = \frac{0.8687}{\sqrt{3}} \approx 0.5015 m \approx 50 cm$ .

✓Verify

Checking units: $x = 0.50 m = 50 cm$ . The horizontal segment in air is $0.5 \sqrt{3} \approx 0.866 m$ , segment in water is $1.5 \times 0.854 \approx 1.281 m$ . Sum $= 0.866 + 1.281 = 2.147 \approx 2.15 m$ . Perfectly matches.

Hints that build this answer step by step

What is the angle of incidence $i$ of the sunlight at the air-water interface?
$i = 60^{\circ}$
Using Snell's Law $1 \cdot \sin (60^{\circ}) = \frac{4}{3} \sin (r)$ , what is the horizontal displacement (shadow cast underwater) due to the submerged depth of $1.5 m$ ?
$1.5 \tan (r) \approx 1.284 m$
If the total underwater shadow is $2.15 m$ , what is the height of the pole above the water surface, $h_{a}$ ?
$50 cm$

✓ Source and academic review

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

Quick checks

Students also ask

Why is the angle of incidence

60^{\circ}

and not

30^{\circ}

The angle of incidence is always measured relative to the normal (vertical) to the surface. Since the ray makes $30^{\circ}$ with the horizontal water surface, $i = 90^{\circ} - 30^{\circ} = 60^{\circ}$ .

Answer

Accepted answer: 50 cm

The height of the pole above the water surface is 50 cm.

Why each option works or fails

Step-by-step solution

Given: Refractive index of water $n_{w} = \frac{4}{3}$ , angle of incident sunlight with water surface $θ = 30^{\circ}$ (so angle of incidence $i = 90^{\circ} - 30^{\circ} = 60^{\circ}$ ), depth of pool $d = 1.5 m$ , total length of shadow at the bottom $L = 2.15 m$ .
Find: Height of the pole above the water surface $x$ in centimetres.
Visualise: The top of the pole casts a ray entering water at a horizontal distance $x \tan i = x \tan 60^{\circ} = x \sqrt{3}$ from the pole's vertical line. Inside the water, the ray refracts at angle $r$ to the normal and traverses a depth of $1.5 m$ , adding a horizontal shift of $1.5 \tan r$ . The total shadow length on the bottom is $L = x \sqrt{3} + 1.5 \tan r = 2.15 m$ .
Strategise: Use Snell's law $1 \cdot \sin 60^{\circ} = \frac{4}{3} \sin r$ to find $\sin r$ and $\tan r$ , then solve the linear geometric equation for $x$ .
Execute: From Snell's law: $\sin r = \frac{3 \sqrt{3}}{8} \approx 0.6495$ . Then $\tan r = \frac{\sin r}{\sqrt{1 - \sin^{2} r}} = \frac{3 \sqrt{3}}{\sqrt{64 - 27}} = \frac{3 \sqrt{3}}{\sqrt{37}} \approx \frac{5.196}{6.0827} \approx 0.8542$ . Now solve for $x$ : $x \sqrt{3} + 1.5 (0.8542) = 2.15 ⟹ x \sqrt{3} = 2.15 - 1.2813 = 0.8687 ⟹ x = \frac{0.8687}{\sqrt{3}} \approx 0.5015 m \approx 50 cm$ .
Verify: Checking units: $x = 0.50 m = 50 cm$ . The horizontal segment in air is $0.5 \sqrt{3} \approx 0.866 m$ , segment in water is $1.5 \times 0.854 \approx 1.281 m$ . Sum $= 0.866 + 1.281 = 2.147 \approx 2.15 m$ . Perfectly matches.

Refraction of Light and Shadow Formation at an Air-Water Interface

What feels right?

The height of the pole above the water surface is 50 cm.

Why each option works or fails

Solution

01Given

02Find

03Visualise

04Strategise

05Execute

✓Verify

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