Question

A coin is placed at the bottom of a water tank of depth 0.5 m. The critical angle of water is about 49°.When viewed normally, the depth of the coin appears to be (in m)

Please help I am having an online test today!!!!!
I’ll mark this the brainliest

Answer 1

Answer: 0.5×sin 49⁰

Explanation:

Answer 2

Answer:

the depth of the coin appears to be at $0.5*sin49^{0} m$.

Explanation:

It is given that depth of water tank is $0.5m$

Critical angle is $49^{0}$

Critical angle is defined as the angle of incidence beyond which rays of light passing through a denser medium to the surface of a less dense medium are no longer refracted, instead it is totally reflected.

Mathematically, the relation is

Ratio of apparent depth and actual depth is equal to sine of critical angle.

Let apparent depth be $d$

Hence, putting values we get

$\frac{d}{0.5} =sin49^{0}$

$d=0.5sin49^{0}$

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