A small bulb is placed at the bottom of a tank containing water to a depth of √7 m. The refractive index of water is 4/3. The area of the surface of water through which light from the bulb can emerge out is xπ m^2. The value of x is _____ .

Correct answer: 9. Apply Snell's law, (sin⁡90°)/(sin⁡C)=μ ⇒sinC=1/(μ)⇒R√(R^2+h^2)=3/4 Squaring, 16R^2=9R^2+9h^2 ⇒7R^2=9h^2⇒R=3√7h=3√7×√7=3 m The area of the circle is, A=πR^2=π3^2 m^2⇒α=9

Q88 / 90 · JEE Main 2022 (26 Jun Shift 2) · Physics

A small bulb is placed at the bottom of a tank containing water to a depth of $\sqrt{7} m$ . The refractive index of water is $\frac{4}{3}$ . The area of the surface of water through which light from the bulb can emerge out is $x π m^{2}$ . The value of $x$ is _____ .

Correct answer: 9

Solution

Apply Snell's law, $\frac{\sin 90 °}{\sin C} = μ$

$\Rightarrow \sin C = \frac{1}{μ} \Rightarrow \frac{R}{\sqrt{R^{2} + h^{2}}} = \frac{3}{4}$

Squaring, $16 R^{2} = 9 R^{2} + 9 h^{2}$

$\Rightarrow 7 R^{2} = 9 h^{2} \Rightarrow R = \frac{3}{\sqrt{7}} h = \frac{3}{\sqrt{7}} \times \sqrt{7} = 3 m$

The area of the circle is, $A = π R^{2} = π {(3)}^{2} m^{2} \Rightarrow α = 9$

Hence, the correct answer is 9.

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A small bulb is placed at the bottom of a tank containing water to a depth of 7 m. The refractive index of water is 43. The area of the surface of water through which light from the bulb can emerge out is xπ m2. The value of x is _____ .

Solution

Relevant Questions

A small bulb is placed at the bottom of a tank containing water to a depth of $\sqrt{7} m$ . The refractive index of water is $\frac{4}{3}$ . The area of the surface of water through which light from the bulb can emerge out is $x π m^{2}$ . The value of $x$ is _____ .