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: \[ u - v = 30 \mathrm{~cm} \] \[ u - \frac{u}{3} = 30 \implies u = 45 \mathrm{~cm}, \quad v = 15 \mathrm{~cm} \]
3. Focal length: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{15} + \frac{1}{45} = \frac{4}{60} \] \[ f = \frac{60}{4} = 15 \mathrm{~cm} \] \[ x = 45 \] Therefore, the correct answer is (45).