Question

The co-ordinates of image of point object P formed after two successive reflection in situation as shown in figure considering first reflection at concave mirror and then at convex.

• A. 30 cm, -14 mm
• B. -30 cm, 14 mm
• C. -30 cm, -14 mm
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

Let the object be at $x=-20$ cm, $y=6$ mm. For the concave mirror ($f_1 = -15$ cm): Using the mirror formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$, we have $\frac{1}{v_1} - \frac{1}{-20} = \frac{1}{-15}$. $\frac{1}{v_1} = -\frac{1}{15} - \frac{1}{20} = \frac{-4-3}{60} = -\frac{7}{60}$. $v_1 = -\frac{60}{7}$ cm. Magnification $m_1 = -\frac{v_1}{u_1} = -\frac{-60/7}{-20} = -\frac{3}{7}$. Image height $h_{i1} = m_1 h_o = -\frac{3}{7} \times 6 = -\frac{18}{7}$ mm. The first image is at $(-\frac{60}{7} \text{ cm}, -\frac{18}{7} \text{ mm})$.

For the convex mirror ($f_2 = 20$ cm), the pole is at $x=50$ cm. The object distance $u_2 = 50 - (-\frac{60}{7}) = 50 + \frac{60}{7} = \frac{350+60}{7} = \frac{410}{7}$ cm. Since the object is to the left of the mirror, $u_2 = -\frac{410}{7}$ cm. The object height for the convex mirror is $h_{o2} = -\frac{18}{7}$ mm. Using the mirror formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$: $\frac{1}{v_2} - \frac{1}{-410/7} = \frac{1}{20}$ $\frac{1}{v_2} + \frac{7}{410} = \frac{1}{20}$ $\frac{1}{v_2} = \frac{1}{20} - \frac{7}{410} = \frac{41 - 14}{410} = \frac{27}{410}$. $v_2 = \frac{410}{27}$ cm. The x-coordinate of the final image is $x_{final} = 50 + v_2 = 50 + \frac{410}{27} = \frac{1350+410}{27} = \frac{1760}{27} \approx 65.185$ cm. Magnification $m_2 = -\frac{v_2}{u_2} = -\frac{410/27}{-410/7} = \frac{7}{27}$. The height of the final image is $h_{i2} = m_2 h_{o2} = \frac{7}{27} \times (-\frac{18}{7}) = -\frac{18}{27} = -\frac{2}{3}$ mm. The calculated coordinates $(\frac{1760}{27} \text{ cm}, -\frac{2}{3} \text{ mm})$ do not match any of the given options. Therefore, "None of these" is the correct answer.