1. **Problem statement:** A small lightbulb is suspended $d_1=450$ cm above the water surface, and the water depth is $d_2=180$ cm. The bottom is a perfect mirror. We want to find how far below the mirror surface the image of the bulb appears from above the water, assuming paraxial approximation and refractive indices $n_{air}=1.0$, $n_{water}=1.3$. 2. **Key concepts and formula:** When light passes from air to water, the apparent depth $d_{apparent}$ of an object submerged in water is related to the real depth $d_{real}$ by: $$d_{apparent} = \frac{n_{air}}{n_{water}} d_{real}$$ This formula comes from Snell's law and the paraxial approximation. 3. **Step (a) - Locate the image of the bulb below the mirror:** - The bulb is $d_1=450$ cm above water. - The mirror is at the bottom of the pool, $d_2=180$ cm below water surface. The image of the bulb in the mirror appears as if the bulb is reflected below the mirror by the same distance $d_1 + d_2$ inside the water, but we must account for refraction when viewed from above water. - Distance from water surface to image inside water (real distance): $$d_{image,real} = d_2 + d_2 + d_1 = d_2 + d_2 + d_1$$ Actually, the bulb is above water, so the image formed by the mirror is at depth: $$d_{image,real} = d_2 + d_2 + d_1 = 2 d_2 + d_1$$ But since the bulb is above water, the reflected image is the mirror image of the bulb about the mirror plane, so the image is located $d_1 + 2 d_2$ below the water surface. - To find apparent depth as seen from above water, convert the submerged part $d_2 + d_2 = 2 d_2$ to apparent depth: $$d_{apparent} = d_1 + \frac{n_{air}}{n_{water}} (2 d_2)$$ Plug in values: $$d_{apparent} = 450 + \frac{1.0}{1.3} \times 360 = 450 + 276.92 = 726.92 \text{ cm}$$ So the image appears approximately 727 cm below the water surface. 4. **Step (b) - Effect of 10% variation in refractive index:** - The refractive index varies by 10%, so $n_{water}$ ranges from $1.3 \times 0.9 = 1.17$ to $1.3 \times 1.1 = 1.43$. - Calculate apparent depth for both extremes: $$d_{apparent,min} = 450 + \frac{1.0}{1.43} \times 360 = 450 + 251.75 = 701.75$$ $$d_{apparent,max} = 450 + \frac{1.0}{1.17} \times 360 = 450 + 307.69 = 757.69$$ - The spread in apparent position is: $$\Delta d = 757.69 - 701.75 = 55.94 \text{ cm}$$ **Final answers:** - (a) The image appears approximately 727 cm below the water surface. - (b) The apparent vertical position spreads by about 56 cm due to refractive index variation.