Question

A fish is 11.9 cm from the front surface of a fish bowl of radius 33 cm. Where does the fish appear to be to someone in air viewing it from in front of the bowl

Answer 1

Answer:

The fish would appear 42.7 cm on the left side from the front of the bowl.

Explanation:

The fish (object) distance = 11.9 cm, radius of curvature of the bowl = 33 cm. The distance of image of the fish (image distance) can be determined by applying the mirror formula;

$\frac{1}{f}$ = $\frac{1}{u}$ + $\frac{1}{v}$

where f is the focal length of the reflecting surface, u is the object distance and v is the image distance.

But, f = $\frac{radius of curvature}{2}$

         = $\frac{33}{2}$

       f = 16.5 cm

Substitute f = 16.5 = $\frac{165}{10}$, and u = 11.9 = $\frac{119}{10}$ in equation 1;

$\frac{10}{165}$ = $\frac{10}{119}$ + $\frac{1}{v}$

$\frac{1}{v}$ = $\frac{10}{165}$ - $\frac{10}{119}$

  = $\frac{1190 - 1650}{19635}$

$\frac{1}{v}$ = $\frac{-460}{19635}$

⇒ v  = $\frac{19635}{-460}$

       = -42.6848

    v = 42.7 cm

The fish would appear 42.7 cm on the left side from the front of the bowl.