Question

a real inverted image in a concave mirror is represented by graph (u ,v , f are coordinates)

Answer 1

The graph is a hyperbolic curve in the third quadrant (u<0, v<0) with asymptotes at u=f and v=f, where f is the negative focal length of the concave mirror.

Answer 2

To represent a real, inverted image formed by a concave mirror using a graph of object distance ($u$) versus image distance ($v$), we use the mirror formula and sign conventions.

1. Mirror Formula: The relationship between object distance ($u$), image distance ($v$), and focal length ($f$) for a spherical mirror is given by: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

2. Sign Conventions (New Cartesian Sign Convention):

• Concave Mirror: The focal length ($f$) is always negative. So, $f < 0$.
• Real Object: The object is always placed to the left of the mirror, so the object distance ($u$) is negative. So, $u < 0$.
• Real Image: A real image is formed in front of the mirror (to the left), so the image distance ($v$) is negative. So, $v < 0$.

3. Conditions for Real and Inverted Image: For a concave mirror, a real and inverted image is formed when the object is placed beyond the principal focus (F), i.e., $u < f$ (since $f$ is negative, this means $|u| > |f|$). In this case, both $u$ and $v$ are negative. Therefore, the graph representing a real, inverted image will lie entirely in the third quadrant of the $u-v$ plane.

4. Analyzing the Graph: From the mirror formula, we can express $v$ in terms of $u$ and $f$: $v = \frac{fu}{u-f}$ Let $f = -F_0$, where $F_0$ is the magnitude of the focal length ($F_0 > 0$). $v = \frac{-F_0 u}{u - (-F_0)} = \frac{-F_0 u}{u + F_0}$ This equation represents a rectangular hyperbola.

• Asymptotes:

  • A vertical asymptote occurs when the denominator is zero: $u + F_0 = 0 \implies u = -F_0$. Since $f = -F_0$, the vertical asymptote is $u = f$.
  • A horizontal asymptote occurs as $u \to \pm \infty$: $v = \lim_{u \to \pm \infty} \frac{-F_0 u}{u + F_0} = \lim_{u \to \pm \infty} \frac{-F_0}{1 + F_0/u} = -F_0$. So, the horizontal asymptote is $v = f$.

• Behavior for Real Image ($u < 0, v < 0$): For the image to be real ($v < 0$), the denominator ($u+F_0$) must be negative, as the numerator ($-F_0 u$) is positive (since $F_0 > 0$ and $u < 0$). So, $u + F_0 < 0 \implies u < -F_0$. This means the object must be placed beyond the focal point (F).

  Let's examine the curve in this region ($u < -F_0$):

  • As $u \to -F_0^-$ (object approaching F from the left, i.e., from beyond C), $u+F_0 \to 0^-$. Then $v \to \frac{\text{positive}}{\text{small negative}} \to -\infty$.
  • As $u \to -\infty$ (object at infinity), $v \to -F_0$ (i.e., $v \to f$).
  • A significant point on this curve is when the object is at the center of curvature ($u = 2f = -2F_0$). In this case, the image is also formed at the center of curvature ($v = 2f = -2F_0$). So, the point $(2f, 2f)$ lies on the graph.

Conclusion: The graph representing a real inverted image in a concave mirror is a hyperbolic branch located in the third quadrant ($u < 0, v < 0$). This branch has asymptotes at $u = f$ and $v = f$ (where $f$ is the negative focal length of the concave mirror). The curve extends from $(f, -\infty)$ as $u$ approaches $f$ from the left, and approaches $(-\infty, f)$ as $u$ tends to $-\infty$. It passes through the point $(2f, 2f)$.