Question

A light source placed at the bottom of a water beaker 10 cm deep forms an illuminated circle of radius 11.3 cm at its surface. If the depth of the water in the beaker is increased to 20 cm,the radius of the illuminated circle will be(${\mu }_w=\frac{4}{3}$):

• A. $16.8cm$
• B. $30.6cm$
• C. $22.6cm$
• D. $33.9cm$

Answer 1

Answer: (C)

$22.6cm$

$\frac{r}{\sqrt{d^2+r^2}}=\frac{1}{\mu }$

$\frac{r_1}{\sqrt{{d_1}^2+{r_1}^2}}=\frac{r_2}{\sqrt{{d_2}^2+{r_2}^2}}$

$\frac{{r_1}^2}{{d_1}^2+{r_1}^2}=\frac{{r_2}^2}{{d_2}^2+{r_2}^2}$

$\frac{{\left(11.3\right)}^2}{227.69}=\frac{{r_2}^2}{400+{r_2}^2}$

$\Rightarrow 51076=100r^2$

$r=\sqrt{\frac{51076}{100}}=22.6cm$