A rod of length $10 ext{ cm}$ lies along the principal axis of a concave mirror of focal length $10 ext{ cm}

Question

A rod of length $10	ext{ cm}$ lies along the principal axis of a concave mirror of focal length $10	ext{ cm}$ in such a way that its end closer to the pole is $20	ext{ cm}$ away from the mirror. The length of the image is:

Accepted Answer

1.  **Given Data:**
    *   Focal length of the concave mirror, $f = -10	ext{ cm}$ (using sign convention).
    *   Length of the rod $= 10	ext{ cm}$.
    *   Distance of the nearer end from the pole, $u_1 = -20	ext{ cm}$.
    *   Distance of the farther end from the pole, $u_2 = -(20 + 10)	ext{ cm} = -30	ext{ cm}$.

2.  **Image of the Nearer End ($v_1$):**
    *   Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
    *   $\frac{1}{v_1} + \frac{1}{-20} = \frac{1}{-10}$
    *   $\frac{1}{v_1} = \frac{1}{20} - \frac{1}{10} = \frac{1 - 2}{20} = -\frac{1}{20}$
    *   $v_1 = -20	ext{ cm}$.
    *   *(Note: The object is at the center of curvature $C = 2f = 20	ext{ cm}$, so the image is also formed at $C$.)*

3.  **Image of the Farther End ($v_2$):**
    *   Using the mirror formula for $u_2 = -30	ext{ cm}$:
    *   $\frac{1}{v_2} + \frac{1}{-30} = \frac{1}{-10}$
    *   $\frac{1}{v_2} = \frac{1}{30} - \frac{1}{10} = \frac{1 - 3}{30} = -\frac{2}{30} = -\frac{1}{15}$
    *   $v_2 = -15	ext{ cm}$.

4.  **Length of the Image:**
    *   The length of the image is the distance between the images of the two ends.
    *   $L_{image} = |v_1 - v_2| = |-20 - (-15)| = |-5| = 5	ext{ cm}$.

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