Analysis of Visual Angles
The problem states that the person sees their image in the silvered ball (a convex mirror) and in a plane mirror with the same angular size. Let the height of the person be $h_0$.
1. Angular size in Plane Mirror:
For a plane mirror at distance $d$, the virtual image forms at a distance $d$ behind the mirror. Thus, the total distance between the eye and the image is $2d$. The angular size $\alpha$ is given by:
$$ \alpha \approx \frac{h_0}{2d} $$
2. Angular size in Convex Mirror (Ball):
Let $u$ be the distance from the person’s eye to the center of the ball. From the geometry shown in the diagram:
$$ u = \sqrt{s_0^2 + (h_0 – h)^2} $$
Given $h_0 = 2.0\,\text{m}$ and $h = 1.0\,\text{m}$, we have $(h_0 – h) = 1.0\,\text{m} = \frac{h_0}{2}$. Thus:
$$ u^2 = s_0^2 + \left(\frac{h_0}{2}\right)^2 = s_0^2 + \frac{h_0^2}{4} = \frac{4s_0^2 + h_0^2}{4} $$
For the convex mirror of radius $r$, the focal length is $f = r/2$. Since the distance $u$ ($>2$m) is much larger than the radius $r$ ($5$cm), the image forms very close to the focus.
Using the mirror equation $\frac{1}{v} + \frac{1}{u} = \frac{2}{r}$ and approximating for $u \gg r$:
$$ v \approx \frac{r}{2} $$
The magnification $m$ is:
$$ m = \frac{v}{u} \approx \frac{r/2}{u} = \frac{r}{2u} $$
The size of the image $h_i$ is:
$$ h_i = m \cdot h_0 = \frac{h_0 r}{2u} $$
The distance of this image from the eye is approximately $u$ (since $v$ is negligible compared to $u$). The angular size $\beta$ is:
$$ \beta \approx \frac{h_i}{u} = \frac{h_0 r}{2u^2} $$
3. Equating the Angular Sizes:
Since the person sees the images as having the same size ($\alpha = \beta$):
$$ \frac{h_0}{2d} = \frac{h_0 r}{2u^2} $$
$$ \frac{1}{d} = \frac{r}{u^2} \implies d = \frac{u^2}{r} $$
Substituting the expression for $u^2$:
$$ d = \frac{1}{r} \left( \frac{4s_0^2 + h_0^2}{4} \right) = \frac{4s_0^2 + h_0^2}{4r} $$
Calculation
Substituting the given values:
- $s_0 = 2.0$ m
- $h_0 = 2.0$ m
- $r = 5.0$ cm $= 0.05$ m