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JEE Physics 2023 Question with Solution

A pole is vertically submerged in a swimming pool, such that it gives a length of shadow 2.15 m2.15 \, \text{m} within water when sunlight is incident at an angle of 30∘30^\circ with the surface of water. If the swimming pool is filled to a height of 1.5 m1.5 \, \text{m}, then the height of the pole above the water surface in centimeters is (nw=43n_w = \frac{4}{3}):

Answer

Correct answer:50

Step-by-step solution

Standard Method

Given: The shadow length within water is 2.15 m2.15 \, \text{m}, the water depth is 1.5 m1.5 \, \text{m}, the sunlight is incident at 30∘30^\circ with the surface of water, and refractive index of water is 43\frac{4}{3}.

Find: The height of the pole above the water surface.

Ray diagram of a vertical pole partly above water, with incident sunlight, normal at water surface, angles 60 degree and r, water depth 1.5 m, underwater shadow x, and total shadow 2.15 m.

Since the angle with the surface is 30∘30^\circ, the angle of incidence with the normal is 60∘60^\circ. Using Snell's law,

sin⁡60∘=43sin⁡r\sin 60^\circ = \frac{4}{3} \sin r

So,

sin⁡r=34×32=338\sin r = \frac{3}{4} \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{8}

Also,

cos⁡r=1−2764=3764=378\cos r = \sqrt{1 - \frac{27}{64}} = \sqrt{\frac{37}{64}} = \frac{\sqrt{37}}{8}

Hence,

tan⁡r=2737≈0.85\tan r = \sqrt{\frac{27}{37}} \approx 0.85

Let the underwater horizontal part of the shadow be xx. Then,

x1.5=0.85\frac{x}{1.5} = 0.85

Therefore,

x=0.85×1.5=1.275 mx = 0.85 \times 1.5 = 1.275 \, \text{m}

The remaining horizontal distance in air is

2.15−1.275=0.875 m2.15 - 1.275 = 0.875 \, \text{m}

Let the height of the pole above water be yy. Then,

tan⁡30∘=y0.875\tan 30^\circ = \frac{y}{0.875}

So,

y=0.875×13=0.50 my = 0.875 \times \frac{1}{\sqrt{3}} = 0.50 \, \text{m}

Thus,

y=50 cmy = 50 \, \text{cm}

Therefore, the height of the pole above the water surface is 50 cm50 \, \text{cm}.

Snell’s Law and Pole Height Calculation

Given: Air to water refraction, incident angle with surface 30∘30^\circ, water depth 1.5 m1.5 \, \text{m}, total shadow length 2.15 m2.15 \, \text{m}, and nw=43n_w = \frac{4}{3}.

Find: Height of the pole above the water surface.

Using Snell's law,

n1sin⁡i=n2sin⁡rn_1 \sin i = n_2 \sin r

Here,

n1=1,i=60∘,n2=43n_1 = 1, \quad i = 60^\circ, \quad n_2 = \frac{4}{3}

Thus,

sin⁡60∘=43sin⁡r\sin 60^\circ = \frac{4}{3} \sin r

which gives

sin⁡r=34sin⁡60∘=34×32=338\sin r = \frac{3}{4} \sin 60^\circ = \frac{3}{4} \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{8}

Now,

cos⁡r=1−sin⁡2r=1−(338)2=1−2764=3764=378\cos r = \sqrt{1 - \sin^2 r} = \sqrt{1 - \left(\frac{3\sqrt{3}}{8}\right)^2} = \sqrt{1 - \frac{27}{64}} = \sqrt{\frac{37}{64}} = \frac{\sqrt{37}}{8}

Therefore,

tan⁡r=sin⁡rcos⁡r=33/837/8=3337≈0.85\tan r = \frac{\sin r}{\cos r} = \frac{3\sqrt{3}/8}{\sqrt{37}/8} = \frac{3\sqrt{3}}{\sqrt{37}} \approx 0.85

If xx is the underwater horizontal shadow length, then

tan⁡r=x1.5\tan r = \frac{x}{1.5}

So,

x=1.5×0.85=1.275 mx = 1.5 \times 0.85 = 1.275 \, \text{m}

The horizontal distance corresponding to the part above water is

2.15−1.275=0.875 m2.15 - 1.275 = 0.875 \, \text{m}

Let the height above water be yy. Since the ray makes 30∘30^\circ with the water surface,

tan⁡30∘=y0.875\tan 30^\circ = \frac{y}{0.875}

Hence,

y=0.875×tan⁡30∘=0.875×13≈0.50 my = 0.875 \times \tan 30^\circ = 0.875 \times \frac{1}{\sqrt{3}} \approx 0.50 \, \text{m}

Therefore, the height of the pole above the water surface is 50 cm50 \, \text{cm}.

Common mistakes

  • Using 30∘30^\circ directly as the angle of incidence in Snell's law is incorrect because the given angle is with the surface, not with the normal. First convert it to 60∘60^\circ with the normal.

  • Treating the full shadow length 2.15 m2.15 \, \text{m} as formed entirely inside water is incorrect. The shadow has one horizontal part due to the ray path in water and another due to the part of the pole above water.

  • Using tan⁡30∘=0.875y\tan 30^\circ = \frac{0.875}{y} reverses the triangle ratio. For the air part, the vertical side is yy and the horizontal side is 0.875 m0.875 \, \text{m}, so use tan⁡30∘=y0.875\tan 30^\circ = \frac{y}{0.875}.

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