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 x π m 2 . The value of x is _____ .

Apply Snell's law, sin ⁡ 90 ° sin ⁡ C = μ ⇒ sin C = 1 μ ⇒ R R 2 + h 2 = 3 4 Squaring, 16 R 2 = 9 R 2 + 9 h 2 ⇒ 7 R 2 = 9 h 2 ⇒ R = 3 7 h = 3 7 × 7 = 3 m The area of the circle is, A = π R 2 = π 3 2 m 2 ⇒ α = 9

Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsRefraction at plane surfaceHard2 minPYQ_2022

PhysicsHardnumerical

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 _____ .

Answer:

9.00

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, $A = π R^{2} = π {(3)}^{2} m^{2} \Rightarrow α = 9$

Stream:JEESubject:PhysicsTopic:Geometrical OpticsSubtopic:Refraction at plane surface

⏱ 2mℹ️ Source: PYQ_2022

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