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a small lightbulb suspended at distance 250 cm above the surface of the water in a swimming pool where the water depth is 200 cm. The bottom of the pool is a large mirror. How far below the mirror surface is the image of the bulb?

Dear Student,

See the diagram.

The apparent shift of the mirror within the water is, h = 200 (1 – 1/μ)

μ = 1.33 is the refractive index of water.

So, the apparent distance between the mirror and the bulb is = 450 – [200 (1 – 1/μ)] = 400.38 cm

The image is also formed at a distance 400.38 cm below this apparent position of the mirror.

Therefore, the actual distance between the mirror and the image is

= (400.38 - h) = 400.38 - [200 (1 – 1/μ)] = 350.76 cm

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i think its 250

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 i tink 250 cm + 200cm =450cm

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I think, as refractive index of water is 1.33, the effective depth of water will be somewhere around 150 cm(200/1.33)

Therefore, the virtual image will be formed at a distance of 400m(250+150) from the surface of the mirror.

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I think, as the refractive index of water is 1.3, the effective depth of water will be somewhere around 150 cm. therefore, the image will be formed at the distance of 400 cm.

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