at distance d, = 320 cm above the surface of the water in a swimming pool where the pool is a large mirror. How far below the mirror surface is the image of the bulb? (Hint: ure (b) below, but take into account the bending of light rays by refraction at the water of the air and the water are 1.00 and 1.33 respectively, and the rays are close to a ll-angle approximation in which sin e tan e= 0.)

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The figure shows a small light bulb suspended at distance \( d_1 = 320 \, \text{cm} \) above the surface of the water in a swimming pool where the water depth is \( d_2 = 270 \, \text{cm} \). The bottom of the pool is a large mirror. How far below the mirror surface is the image of the bulb? (Hint: Construct a diagram of two rays like that of figure (b) below, but take into account the bending of light rays by refraction at the water surface. Assume that the indices of refraction of the air and the water are 1.00 and 1.33 respectively, and the rays are close to a vertical axis through the bulb, and use the small-angle approximation in which \( \sin \theta \approx \tan \theta \approx \theta \).

**Diagram (a):**
- The diagram shows the setup with the light bulb, the water surface, and the bottom mirror.
- The light bulb is at height \( d_1 = 320 \, \text{cm} \) above the water.
- The water depth is \( d_2 = 270 \, \text{cm} \).

**Diagram (b):**
- It illustrates the path of the light rays from the bulb to the mirror and back.
- The rays refract at the water surface due to the change in the medium from air to water.
- \( \theta \) represents the angle of refraction.
- The distance where the image forms below the mirror is related to the angles and distances indicated.

**Question:**
Calculate the distance below the mirror where the image of the bulb forms by considering the refraction and using the given indices of refraction. Enter the result in the fields below.

Number: [Input field]
Units: [Dropdown]
Transcribed Image Text:The figure shows a small light bulb suspended at distance \( d_1 = 320 \, \text{cm} \) above the surface of the water in a swimming pool where the water depth is \( d_2 = 270 \, \text{cm} \). The bottom of the pool is a large mirror. How far below the mirror surface is the image of the bulb? (Hint: Construct a diagram of two rays like that of figure (b) below, but take into account the bending of light rays by refraction at the water surface. Assume that the indices of refraction of the air and the water are 1.00 and 1.33 respectively, and the rays are close to a vertical axis through the bulb, and use the small-angle approximation in which \( \sin \theta \approx \tan \theta \approx \theta \). **Diagram (a):** - The diagram shows the setup with the light bulb, the water surface, and the bottom mirror. - The light bulb is at height \( d_1 = 320 \, \text{cm} \) above the water. - The water depth is \( d_2 = 270 \, \text{cm} \). **Diagram (b):** - It illustrates the path of the light rays from the bulb to the mirror and back. - The rays refract at the water surface due to the change in the medium from air to water. - \( \theta \) represents the angle of refraction. - The distance where the image forms below the mirror is related to the angles and distances indicated. **Question:** Calculate the distance below the mirror where the image of the bulb forms by considering the refraction and using the given indices of refraction. Enter the result in the fields below. Number: [Input field] Units: [Dropdown]
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