Question

Distance between object and its image (magnified by $-\frac{1}{3}$ ) is 30 cm. The focal length of the mirror used is $\left(\frac{\mathrm{x}}{4}\right) \mathrm{cm}$, where magnitude of value of x is _______ .

Hint:

Use the magnification formula to find the object and image distances, then calculate the focal length.

Answer 1

Correct Answer: 45

1. Given: \[ m = -\frac{1}{3} \] \[ -\frac{v}{u} = \frac{1}{3} \implies v = -\frac{u}{3} \]
2. Distance between object and image: \[ u - v = 30 \mathrm{~cm} \] \[ u - (-\frac{u}{3}) = 30 \implies u + \frac{u}{3} = 30 \implies \frac{4u}{3} = 30 \implies u = \frac{90}{4} = 22.5 \mathrm{~cm}, \quad v = -\frac{22.5}{3} = -7.5 \mathrm{~cm} \]
3. Focal length: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-7.5} + \frac{1}{22.5} = \frac{-3+1}{22.5} = \frac{-2}{22.5} \] \[ f = \frac{22.5}{-2} = -11.25 \mathrm{~cm} \] The object distance is 22.5 cm.Therefore, the correct answer is (22.5).