Question

The figure shows a small lightbulb suspended
at distance $$d_1=250$$ cm above the
surface of the water in a swimming
pool where the water depth is $$d_2 = 200$$ cm. The bottom of the pool is a
large mirror. How far below the mirror surface is the image of the bulb? (Hint: Assume that the rays are
close to a vertical axis through the bulb, and use the small-angle approximation in which $$\sin \theta \approx \tan \theta \approx \theta$$.)

Answer 1

We apply the law of refraction, assuming all angles are in radians:
$$\dfrac{\sin \theta}{\sin \theta'}=\dfrac{n_w}{n_{air}}$$,
which in our case reduces to $$θ' ≈ θ/n_w$$ (since both $$θ$$ and $$θ'$$ are small, and $$n_{air} ≈ 1$$). We refer to our figure on the right.
The object O is a vertical distance $$d_1$$ above the water, and
the water surface is a vertical distance $$d_2$$ above the mirror.
We are looking for a distance d (treated as a positive
number) below the mirror where the image I of the object is
formed. In the triangle O AB
$$|AB|=d_1\tan\theta\approx d_1\theta$$
and in the triangle CBD
$$|BC|=2d_2\tan\theta'\approx 2d_2\theta'\approx \dfrac{2d_2\theta}{n_w}$$
Finally, in the triangle ACI, we have $$|AI| = d + d_2$$. Therefore,
$$d=|AI|-d_2=\dfrac{|AC|}{\tan\theta}-d_2\,\approx\dfrac{|AB|+|BC|}{\theta}-d_2=\Bigg(d_1\theta+\dfrac{2d_2\theta}{n_w}\Bigg)\dfrac{1}{\theta}-d_2=d_1+\dfrac{2d_2}{n_w}-d_2$$
$$=250\,cm+\dfrac{2(200\,cm)}{1.3 3}-200\,cm=351\,cm$$