Example 9.3 An object is placed at (i) 10 cm, (ii) 5 cm in front of a concave mirror of radius of curvature 15 cm. Find the position, nature, and magnification of the image in each case.

Step-by-step solution

1. 1

   Given values

   Concave mirror, $R = 15 \text{ cm}$. Object distances: (i) $u_1 = -10 \text{ cm}$, (ii) $u_2 = -5 \text{ cm}$.

   $$R = 15 \text{ cm}, u_1 = -10 \text{ cm}, u_2 = -5 \text{ cm}$$
2. 2

   Sign convention & Focal length

   For concave mirror, $R$ is negative. Focal length $f = R/2$.

   $$f = R/2$$
3. 3

   Mirror & Magnification Formula

   Use mirror formula $1/f = 1/v + 1/u$ and magnification $m = -v/u$.

   $$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}, m = -\frac{v}{u}$$
4. 4

   Calculate focal length

   Focal length $f = -15/2 = -7.5 \text{ cm}$.

   $$f = -7.5 \text{ cm}$$
5. 5

   Case (i): $u_1 = -10 \text{ cm}$

   Using mirror formula: $1/v_1 = 1/f - 1/u_1 = 1/(-7.5) - 1/(-10)$.

   $$\frac{1}{v_1} = \frac{1}{-7.5} - \frac{1}{-10}$$
6. 6

   Solve for $v_1$

   $1/v_1 = -2/15 + 1/10 = (-4+3)/30 = -1/30$. So $v_1 = -30 \text{ cm}$.

   $$v_1 = -30 \text{ cm}$$
7. 7

   Magnification $m_1$

   $m_1 = -v_1/u_1 = -(-30)/(-10) = -30/10 = -2$.

   $$m_1 = -\frac{-30}{-10} = -2$$
8. 8

   Nature of image (i)

   $v_1 < 0$ (real), $m_1 < 0$ (inverted), $|m_1| > 1$ (magnified).
9. 9

   Case (ii): $u_2 = -5 \text{ cm}$

   Using mirror formula: $1/v_2 = 1/f - 1/u_2 = 1/(-7.5) - 1/(-5)$.

   $$\frac{1}{v_2} = \frac{1}{-7.5} - \frac{1}{-5}$$
10. 10

    Solve for $v_2$

    $1/v_2 = -2/15 + 1/5 = (-2+3)/15 = 1/15$. So $v_2 = 15 \text{ cm}$.

    $$v_2 = 15 \text{ cm}$$