## Related questions with answers

A narrow beam of light from a laser travels through air (n=1.00) and strikes point A on the surface of the water (n=1.33) in a lake. The angle of incidence is $55^{\circ}$. The depth of the lake is 3.0 m. On the flat lake-bottom is point B, directly below point A. (a) If refraction did not occur, how far away from point B would the laser beam strike the lake-bottom? (b) Considering refraction, how far away from point B would the laser beam strike the lake-bottom?

Solutions

Verified**Given**
The index of refraction of air is $n_1=1$.
The index of refraction of water is $n_2=1.33$.
The angle of incidence is $\theta_1=55\degree$.
The depth of the lake is $d=3\ \text{m}$.

We are given the angle of incidence of the light to be $\theta_i=55 \text{\textdegree}$. If (a) refraction does not occur, the angle of the light continues to be $55\text{\textdegree}$ in the water (relative to the normal). Thus, $\tan 55\text{\textdegree} = \frac{d}{3.0m} \rightarrow d=4.3m$.

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